💡Okay, so I have this inequality to prove: for nonnegative real numbers (a, b, c), the expression (a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a)) is greater than or equal to zero. Hmm, that looks a bit complicated, but let me try to break it down.First, I should probably expand each term to see if I can simplify the expression. Let me take the first term: (a(a - b)(a - 2b)). If I expand this, it becomes (a(a^2 - 2ab - ab + 2b^2)), which simplifies to (a(a^2 - 3ab + 2b^2)). Multiplying through by (a), I get (a^3 - 3a^2b + 2ab^2).Similarly, expanding the second term (b(b - c)(b - 2c)) gives (b(b^2 - 2bc - bc + 2c^2)), which simplifies to (b^3 - 3b^2c + 2bc^2).And the third term (c(c - a)(c - 2a)) expands to (c(c^2 - 2ac - ac + 2a^2)), simplifying to (c^3 - 3c^2a + 2ca^2).So, putting all these together, the entire expression becomes:[a^3 - 3a^2b + 2ab^2 + b^3 - 3b^2c + 2bc^2 + c^3 - 3c^2a + 2ca^2]Let me rearrange the terms to group similar ones:[(a^3 + b^3 + c^3) + (-3a^2b - 3b^2c - 3c^2a) + (2ab^2 + 2bc^2 + 2ca^2)]Hmm, I notice that each group has a common factor or structure. Maybe I can factor some parts or find a pattern here.Looking at the first group, (a^3 + b^3 + c^3), that's straightforward. The second group is (-3(a^2b + b^2c + c^2a)), and the third group is (2(ab^2 + bc^2 + ca^2)).I wonder if I can factor this expression further or perhaps use some known inequalities. Maybe I can factor by grouping or look for symmetry.Another thought: since the variables are nonnegative, perhaps I can consider specific cases where one variable is zero or where two variables are equal. That might help me understand the behavior of the expression.Let me try setting (c = 0). Then the expression simplifies to:[a(a - b)(a - 2b) + b(b - 0)(b - 0) + 0]Which is:[a(a - b)(a - 2b) + b^3]Expanding (a(a - b)(a - 2b)) as before, we get (a^3 - 3a^2b + 2ab^2). So the entire expression becomes:[a^3 - 3a^2b + 2ab^2 + b^3]Let me factor this. Maybe factor out (a^3 + b^3) first:[a^3 + b^3 - 3a^2b + 2ab^2]I know that (a^3 + b^3 = (a + b)(a^2 - ab + b^2)), but I'm not sure if that helps here. Alternatively, perhaps factor by grouping:[(a^3 - 3a^2b) + (2ab^2 + b^3) = a^2(a - 3b) + b^2(2a + b)]Not sure if that helps. Maybe try plugging in specific values for (a) and (b) to see if the expression is nonnegative.Let me set (a = b). Then the expression becomes:[a(a - a)(a - 2a) + a(a - 0)(a - 0) = 0 + a^3 = a^3 geq 0]Which is true since (a) is nonnegative.What if (a > b)? Let's say (a = 2) and (b = 1). Then the expression is:[2(2 - 1)(2 - 2*1) + 1^3 = 2(1)(0) + 1 = 0 + 1 = 1 geq 0]Okay, that's nonnegative.What if (a < b)? Let's say (a = 1) and (b = 2). Then:[1(1 - 2)(1 - 4) + 8 = 1(-1)(-3) + 8 = 3 + 8 = 11 geq 0]Still nonnegative. Interesting.So, in the case where (c = 0), the expression seems to be nonnegative. Maybe this suggests that the original expression is always nonnegative.Another approach: perhaps consider homogenizing the inequality or using known symmetric inequalities. The expression seems to be symmetric in a cyclic manner, not fully symmetric. Maybe I can use the rearrangement inequality or something similar.Wait, another idea: perhaps express the entire expression as a sum of squares. If I can write it as a sum of squares, then it would be nonnegative.Let me see. The expression is:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Hmm, maybe rearrange terms:[(a^3 - 3a^2b + 2ab^2) + (b^3 - 3b^2c + 2bc^2) + (c^3 - 3c^2a + 2ca^2)]Each of these terms looks similar. Let me focus on one of them, say (a^3 - 3a^2b + 2ab^2). Can I factor this?Yes, factoring:[a(a^2 - 3ab + 2b^2) = a(a - b)(a - 2b)]Which is the original term. So, that doesn't help much.Wait, maybe consider the entire expression as a combination of cyclic terms. Perhaps use the identity for cyclic sums.Alternatively, maybe consider the expression as a determinant or use some substitution. Hmm, not sure.Another idea: perhaps use the AM-GM inequality. But I'm not sure how to apply it here directly.Wait, let me think about the structure of the expression. Each term is of the form (x(x - y)(x - 2y)). Maybe I can analyze each term individually.For each term (x(x - y)(x - 2y)), let's see when it's nonnegative.If (x geq 2y), then (x - y geq 0) and (x - 2y geq 0), so the term is nonnegative.If (y leq x leq 2y), then (x - y geq 0) but (x - 2y leq 0), so the term is nonpositive.If (x leq y), then (x - y leq 0) and (x - 2y leq 0), so the term is nonnegative because negative times negative is positive.So, each individual term can be positive or negative depending on the relationship between (x) and (y). Therefore, the entire sum isn't obviously nonnegative just by looking at individual terms.Hmm, maybe I need to consider the entire expression together rather than individual terms.Another approach: perhaps use the method of Lagrange multipliers or consider taking partial derivatives to find minima, but that might be too complicated.Wait, maybe consider substituting (a = b = c). Then the expression becomes:[a(a - a)(a - 2a) + a(a - a)(a - 2a) + a(a - a)(a - 2a) = 0 + 0 + 0 = 0]So, the expression is zero when all variables are equal. That's a good check.What if two variables are equal? Let's say (a = b). Then the expression becomes:[a(a - a)(a - 2a) + a(a - c)(a - 2c) + c(c - a)(c - 2a)]Simplify:[0 + a(a - c)(a - 2c) + c(c - a)(c - 2a)]Notice that (c(c - a)(c - 2a) = -c(a - c)(2a - c)). So, the expression becomes:[a(a - c)(a - 2c) - c(a - c)(2a - c)]Factor out ((a - c)):[(a - c)[a(a - 2c) - c(2a - c)] = (a - c)[a^2 - 2ac - 2ac + c^2] = (a - c)(a^2 - 4ac + c^2)]Hmm, that's ((a - c)(a^2 - 4ac + c^2)). Let me see if this is nonnegative.Let me denote (d = a - c). Then the expression is (d(a^2 - 4ac + c^2)). Let me express (a^2 - 4ac + c^2) in terms of (d). Since (d = a - c), we have (a = c + d). Substitute into (a^2 - 4ac + c^2):[(c + d)^2 - 4(c + d)c + c^2 = c^2 + 2cd + d^2 - 4c^2 - 4cd + c^2 = (-2c^2 - 2cd + d^2)]So, the expression becomes:[d(-2c^2 - 2cd + d^2)]Hmm, not sure if this helps. Maybe consider specific values again.Let me set (a = b = 1) and vary (c). Then the expression becomes:[1(1 - 1)(1 - 2*1) + 1(1 - c)(1 - 2c) + c(c - 1)(c - 2*1)]Simplify:[0 + (1 - c)(1 - 2c) + c(c - 1)(c - 2)]Expand each term:First term: ((1 - c)(1 - 2c) = 1 - 2c - c + 2c^2 = 1 - 3c + 2c^2)Second term: (c(c - 1)(c - 2) = c(c^2 - 3c + 2) = c^3 - 3c^2 + 2c)So, total expression:[1 - 3c + 2c^2 + c^3 - 3c^2 + 2c = c^3 - c^2 - c + 1]Factor this:[c^3 - c^2 - c + 1 = c^2(c - 1) -1(c - 1) = (c^2 - 1)(c - 1) = (c - 1)(c + 1)(c - 1) = (c - 1)^2(c + 1)]Since (c) is nonnegative, (c + 1 geq 1 > 0), and ((c - 1)^2 geq 0). Therefore, the entire expression is nonnegative. So, when (a = b), the expression is nonnegative.That's a good sign. Maybe the expression is always nonnegative.Another idea: perhaps use the fact that the expression is cyclic and consider the rearrangement inequality or other cyclic inequalities.Wait, another approach: maybe express the entire expression as a determinant of a matrix. Sometimes, expressions can be represented as determinants, which can help in proving nonnegativity.Alternatively, perhaps consider the expression as a combination of squares or other nonnegative terms.Wait, let me try to write the expression as a sum of squares. Let me see:The expression is:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Let me rearrange terms:[(a^3 - 3a^2b + 2ab^2) + (b^3 - 3b^2c + 2bc^2) + (c^3 - 3c^2a + 2ca^2)]Each of these terms is similar. Let me focus on one term, say (a^3 - 3a^2b + 2ab^2). Maybe I can write this as a square or a product.Wait, (a^3 - 3a^2b + 2ab^2 = a(a^2 - 3ab + 2b^2) = a(a - b)(a - 2b)), which is the original term. So, that doesn't help.Alternatively, perhaps complete the square. Let me see:Consider (a^3 - 3a^2b + 2ab^2). Let me factor out (a):[a(a^2 - 3ab + 2b^2) = a(a - b)(a - 2b)]Hmm, same as before.Wait, maybe consider the entire expression as a combination of cyclic terms and use known cyclic inequalities.Alternatively, perhaps use the substitution (x = a - b), (y = b - c), (z = c - a), but I'm not sure.Wait, another idea: perhaps use the identity for (a^3 + b^3 + c^3 - 3abc), which is ((a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)). But our expression is different.Wait, let me compute the expression for specific values to see if it's always nonnegative.Let me try (a = 2), (b = 1), (c = 0). Then the expression becomes:[2(2 - 1)(2 - 2*1) + 1(1 - 0)(1 - 2*0) + 0 = 2(1)(0) + 1(1)(1) + 0 = 0 + 1 + 0 = 1 geq 0]Good.Another test: (a = 1), (b = 2), (c = 3). Then:[1(1 - 2)(1 - 4) + 2(2 - 3)(2 - 6) + 3(3 - 1)(3 - 2)]Compute each term:First term: (1(-1)(-3) = 3)Second term: (2(-1)(-4) = 8)Third term: (3(2)(1) = 6)Total: (3 + 8 + 6 = 17 geq 0)Another test: (a = 3), (b = 1), (c = 2). Then:[3(3 - 1)(3 - 2) + 1(1 - 2)(1 - 4) + 2(2 - 3)(2 - 6)]Compute each term:First term: (3(2)(1) = 6)Second term: (1(-1)(-3) = 3)Third term: (2(-1)(-4) = 8)Total: (6 + 3 + 8 = 17 geq 0)Hmm, seems nonnegative in these cases. Maybe it's always nonnegative.Wait, another test: (a = 1), (b = 0), (c = 0). Then:[1(1 - 0)(1 - 0) + 0 + 0 = 1 geq 0]Good.What about (a = 0), (b = 1), (c = 2). Then:[0 + 1(1 - 2)(1 - 4) + 2(2 - 0)(2 - 0) = 0 + 1(-1)(-3) + 2(2)(2) = 0 + 3 + 8 = 11 geq 0]Still nonnegative.Wait, what if (a = 1), (b = 2), (c = 1). Then:[1(1 - 2)(1 - 4) + 2(2 - 1)(2 - 2) + 1(1 - 1)(1 - 2)]Compute each term:First term: (1(-1)(-3) = 3)Second term: (2(1)(0) = 0)Third term: (1(0)(-1) = 0)Total: (3 + 0 + 0 = 3 geq 0)Another test: (a = 2), (b = 3), (c = 1). Then:[2(2 - 3)(2 - 6) + 3(3 - 1)(3 - 2) + 1(1 - 2)(1 - 4)]Compute each term:First term: (2(-1)(-4) = 8)Second term: (3(2)(1) = 6)Third term: (1(-1)(-3) = 3)Total: (8 + 6 + 3 = 17 geq 0)Hmm, seems consistent. Maybe the expression is always nonnegative.Wait, another idea: perhaps consider the expression as a quadratic in one variable while keeping others fixed. For example, fix (b) and (c), and consider the expression as a function of (a). Then, check if it's always nonnegative.But that might be complicated. Alternatively, perhaps use the fact that the expression is symmetric in a cyclic way and apply some cyclic inequality techniques.Wait, another approach: perhaps use the substitution (x = a - b), (y = b - c), (z = c - a), but I'm not sure.Wait, another idea: perhaps express the entire expression as a product of three terms. Let me see:The original expression is:[a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a)]Let me denote each term as (T_a = a(a - b)(a - 2b)), (T_b = b(b - c)(b - 2c)), (T_c = c(c - a)(c - 2a)).I wonder if (T_a + T_b + T_c) can be factored or expressed in terms of symmetric polynomials.Alternatively, perhaps consider the expression as a combination of symmetric sums. Let me compute the expression in terms of symmetric sums.Let me recall that symmetric sums are expressions that remain unchanged under permutation of variables. The given expression is cyclic but not symmetric.Wait, perhaps express the expression in terms of elementary symmetric polynomials. Let me denote (S_1 = a + b + c), (S_2 = ab + bc + ca), (S_3 = abc).But I'm not sure if that helps directly. Let me try expanding the entire expression again:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Let me rearrange the terms:[(a^3 + b^3 + c^3) + (-3a^2b - 3b^2c - 3c^2a) + (2ab^2 + 2bc^2 + 2ca^2)]Hmm, maybe factor out common terms:[(a^3 + b^3 + c^3) - 3(a^2b + b^2c + c^2a) + 2(ab^2 + bc^2 + ca^2)]Let me see if I can write this as a combination of known identities.I recall that (a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)). But our expression is different.Wait, let me compute the difference between our expression and (a^3 + b^3 + c^3 - 3abc):Our expression: (a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2)Minus (a^3 + b^3 + c^3 - 3abc): (-3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2 + 3abc)Hmm, not sure if that helps.Wait, another idea: perhaps express the expression as a sum involving ((a - b)^2), which are always nonnegative.Let me try to write the expression as a sum of squares. Let me see:Suppose I can write the expression as:[sum (a - b)^2 cdot text{something}]But I'm not sure what the "something" would be.Alternatively, perhaps consider the expression as a determinant. Let me try:The expression is:[a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a)]Let me write each term as (x(x - y)(x - 2y)). Maybe this can be represented as a determinant.Wait, another idea: perhaps use the identity for (x(x - y)(x - 2y)). Let me compute it:(x(x - y)(x - 2y) = x(x^2 - 3xy + 2y^2) = x^3 - 3x^2y + 2xy^2)So, the entire expression is:[a^3 - 3a^2b + 2ab^2 + b^3 - 3b^2c + 2bc^2 + c^3 - 3c^2a + 2ca^2]Which is the same as before.Wait, maybe consider the expression as a combination of cyclic permutations. Let me see:Let me write the expression as:[sum_{cyc} a^3 - 3sum_{cyc} a^2b + 2sum_{cyc} ab^2]Which is:[sum_{cyc} a^3 - 3sum_{cyc} a^2b + 2sum_{cyc} ab^2]Hmm, maybe factor this expression.Wait, another idea: perhaps factor out (a - b), (b - c), etc., but I'm not sure.Wait, another approach: perhaps use the substitution (x = a - b), (y = b - c), (z = c - a), but that might complicate things.Wait, another idea: perhaps express the expression in terms of (a + b + c), (ab + bc + ca), and (abc). Let me try.Let me denote (S = a + b + c), (P = ab + bc + ca), (Q = abc).But I'm not sure how to express the given expression in terms of (S), (P), and (Q). Let me try expanding the expression:The expression is:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Let me note that (a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b + c)(ab + bc + ca) + 3abc). So:[a^3 + b^3 + c^3 = S^3 - 3SP + 3Q]Similarly, (a^2b + b^2c + c^2a) can be expressed as:[a^2b + b^2c + c^2a = (a + b + c)(ab + bc + ca) - (a^2c + b^2a + c^2b) - 3abc]Wait, that seems complicated. Maybe another identity.I recall that (a^2b + b^2c + c^2a + a^2c + b^2a + c^2b = (a + b + c)(ab + bc + ca) - 3abc). So:[a^2b + b^2c + c^2a + a^2c + b^2a + c^2b = SP - 3Q]Therefore, (a^2b + b^2c + c^2a = frac{SP - 3Q - (a^2c + b^2a + c^2b)}{1}). Hmm, not helpful.Wait, perhaps express (a^2b + b^2c + c^2a) in terms of symmetric sums. Let me see:I know that (a^2b + b^2c + c^2a) is a cyclic sum, not symmetric. Similarly, (ab^2 + bc^2 + ca^2) is another cyclic sum.Let me denote (C = a^2b + b^2c + c^2a) and (D = ab^2 + bc^2 + ca^2). Then, the expression becomes:[(a^3 + b^3 + c^3) - 3C + 2D]I also know that (a^3 + b^3 + c^3 = S^3 - 3SP + 3Q), as before.So, substituting:[(S^3 - 3SP + 3Q) - 3C + 2D]But I still need to express (C) and (D) in terms of (S), (P), and (Q). I'm not sure if that's possible.Wait, another idea: perhaps consider the difference between (C) and (D). Let me compute (C - D):[(a^2b + b^2c + c^2a) - (ab^2 + bc^2 + ca^2) = a^2b - ab^2 + b^2c - bc^2 + c^2a - ca^2]Factor each pair:[ab(a - b) + bc(b - c) + ca(c - a)]Hmm, interesting. So, (C - D = ab(a - b) + bc(b - c) + ca(c - a)). Not sure if that helps.Wait, another idea: perhaps express the entire expression as a product of three terms. Let me see:The original expression is:[a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a)]Let me denote (x = a - b), (y = b - c), (z = c - a). Then, (x + y + z = 0). Maybe this substitution can help.But I'm not sure. Let me try expressing each term in terms of (x), (y), (z):First term: (a(a - b)(a - 2b) = a x (a - 2b)). But (a - 2b = (a - b) - b = x - b). Hmm, not helpful.Wait, another idea: perhaps use the fact that the expression is cyclic and apply the rearrangement inequality. The rearrangement inequality states that for two sequences ordered in the same way, the sum of their products is maximized, and if ordered oppositely, minimized.But I'm not sure how to apply it here directly.Wait, another approach: perhaps consider the expression as a quadratic form. Let me see:The expression is:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Let me try to write this as a quadratic in (a), treating (b) and (c) as constants.So, treating (a) as variable:[a^3 - 3a^2b + 2ab^2 + (b^3 + c^3 - 3b^2c - 3c^2a + 2bc^2 + 2ca^2)]Wait, that seems messy. Maybe not helpful.Wait, another idea: perhaps use the identity for (a^3 + b^3 + c^3 - 3abc), which is ((a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)). Let me see if I can relate our expression to this.Our expression is:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Let me subtract (3abc) from both sides:[(a^3 + b^3 + c^3 - 3abc) - 3(a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2)]So, the expression becomes:[(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) - 3(a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2)]Hmm, not sure if that helps. Maybe factor further.Wait, another idea: perhaps express the expression as a determinant of a matrix. Let me try:Consider the matrix:[begin{pmatrix}a & b & c c & a & b b & c & aend{pmatrix}]The determinant of this matrix is (a^3 + b^3 + c^3 - 3abc), which is similar to part of our expression. But our expression is different.Wait, another idea: perhaps use the identity for the sum of cubes. Let me recall that (a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)). So, if I can express our expression in terms of this identity, maybe I can proceed.Our expression is:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Let me write this as:[(a^3 + b^3 + c^3 - 3abc) - 3(a^2b + b^2c + c^2a - ab^2 - bc^2 - ca^2) + 3abc]Wait, let me check:Original expression:[a^3 + b^3 + c^3 - 3a^2b - 3b^2c - 3c^2a + 2ab^2 + 2bc^2 + 2ca^2]Let me group terms:[(a^3 + b^3 + c^3 - 3abc) + (-3a^2b - 3b^2c - 3c^2a + 3ab^2 + 3bc^2 + 3ca^2) + 3abc - 3abc]Wait, that seems forced. Let me see:[(a^3 + b^3 + c^3 - 3abc) + (-3a^2b - 3b^2c - 3c^2a + 3ab^2 + 3bc^2 + 3ca^2) + 0]So, the expression is:[(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3(ab^2 + bc^2 + ca^2 - a^2b - b^2c - c^2a)]Hmm, not sure if that helps.Wait, another idea: perhaps express the expression as a combination of ((a - b)^2), ((b - c)^2), and ((c - a)^2), which are always nonnegative.Let me try to write the expression as a sum of squares. Let me see:Suppose I can write the expression as:[sum (a - b)^2 cdot text{something}]But I'm not sure what the "something" would be. Let me try expanding ((a - b)^2) and see if it can be related.((a - b)^2 = a^2 - 2ab + b^2). Not directly related.Wait, another idea: perhaps use the identity for the sum of cubes again, but in a different way.Wait, another approach: perhaps use the fact that the expression is cyclic and consider the cases where one variable is the largest, middle, or smallest.Let me assume, without loss of generality, that (a geq b geq c). Then, I can analyze the expression under this ordering.Given (a geq b geq c), let's see the sign of each term:First term: (a(a - b)(a - 2b)). Since (a geq b), (a - b geq 0). Now, (a - 2b) could be positive or negative. If (a geq 2b), then (a - 2b geq 0), so the term is nonnegative. If (b < a < 2b), then (a - 2b < 0), so the term is negative.Second term: (b(b - c)(b - 2c)). Since (b geq c), (b - c geq 0). Similarly, (b - 2c) could be positive or negative. If (b geq 2c), then (b - 2c geq 0), so the term is nonnegative. If (c < b < 2c), then (b - 2c < 0), so the term is negative.Third term: (c(c - a)(c - 2a)). Since (c leq a), (c - a leq 0). Also, (c - 2a leq 0). So, the product (c(c - a)(c - 2a)) is nonnegative because it's the product of three nonpositive terms (since (c geq 0)).Wait, let me clarify:- (c geq 0)- (c - a leq 0) because (a geq c)- (c - 2a leq 0) because (a geq c)So, (c(c - a)(c - 2a) = c cdot (-ve) cdot (-ve) = c cdot (+ve) geq 0)Therefore, the third term is nonnegative.So, under the assumption (a geq b geq c), the third term is always nonnegative. The first and second terms can be positive or negative depending on whether (a geq 2b) or (b geq 2c).Hmm, so maybe the sum of the first and second terms can be negative, but the third term is always nonnegative. I need to see if the third term compensates for any negativity from the first two terms.Alternatively, perhaps consider the entire expression as a combination where the nonnegative third term outweighs any negative contributions from the first two.But I'm not sure. Maybe I need a different approach.Wait, another idea: perhaps use the substitution (x = a - b), (y = b - c), (z = c). Then, (a = z + y + x), (b = z + y), (c = z). Maybe this substitution can help express the expression in terms of (x), (y), (z), which are nonnegative.Let me try:First, express (a), (b), (c) in terms of (x), (y), (z):[a = x + y + z][b = y + z][c = z]Where (x, y, z geq 0).Now, substitute into the expression:First term: (a(a - b)(a - 2b))[= (x + y + z)(x + y + z - (y + z))(x + y + z - 2(y + z))]Simplify:[= (x + y + z)(x)(x - y - z)]Wait, (x - y - z) could be positive or negative. Hmm, not sure.Second term: (b(b - c)(b - 2c))[= (y + z)(y + z - z)(y + z - 2z)]Simplify:[= (y + z)(y)(y - z)]Again, (y - z) could be positive or negative.Third term: (c(c - a)(c - 2a))[= z(z - (x + y + z))(z - 2(x + y + z))]Simplify:[= z(-x - y)(-2x - 2y - z)]Which is:[= z(x + y)(2x + 2y + z)]Since (x, y, z geq 0), this term is nonnegative.So, the third term is nonnegative, but the first two terms can be positive or negative. Hmm.Wait, let me compute each term in terms of (x), (y), (z):First term: ((x + y + z) cdot x cdot (x - y - z))Second term: ((y + z) cdot y cdot (y - z))Third term: (z(x + y)(2x + 2y + z))Let me compute each term:First term:[(x + y + z) cdot x cdot (x - y - z) = x(x + y + z)(x - y - z)]Let me expand this:[x[(x)(x - y - z) + (y + z)(x - y - z)] = x[x^2 - xy - xz + xy - y^2 - yz + xz - yz - z^2]]Simplify:[x[x^2 - y^2 - 2yz - z^2] = x^3 - xy^2 - 2xyz - xz^2]Second term:[(y + z) cdot y cdot (y - z) = y(y + z)(y - z) = y(y^2 - z^2) = y^3 - y z^2]Third term:[z(x + y)(2x + 2y + z) = z[2x(x + y) + z(x + y)] = z[2x^2 + 2xy + xz + yz]]Which is:[2x^2 z + 2x y z + x z^2 + y z^2]Now, sum all three terms:First term: (x^3 - xy^2 - 2xyz - xz^2)Second term: (y^3 - y z^2)Third term: (2x^2 z + 2x y z + x z^2 + y z^2)Adding them together:[x^3 - xy^2 - 2xyz - xz^2 + y^3 - y z^2 + 2x^2 z + 2x y z + x z^2 + y z^2]Simplify term by term:- (x^3)- (-xy^2)- (-2xyz + 2xyz = 0)- (-xz^2 + x z^2 = 0)- (y^3)- (-y z^2 + y z^2 = 0)- (2x^2 z)So, the total expression simplifies to:[x^3 - xy^2 + y^3 + 2x^2 z]Hmm, that's simpler. So, the entire original expression reduces to:[x^3 - xy^2 + y^3 + 2x^2 z]Where (x, y, z geq 0).Now, I need to show that this expression is nonnegative.Let me factor this expression if possible. Let me see:[x^3 - xy^2 + y^3 + 2x^2 z = x^3 + y^3 - xy^2 + 2x^2 z]Hmm, can I factor (x^3 + y^3)? Yes, (x^3 + y^3 = (x + y)(x^2 - xy + y^2)). So:[(x + y)(x^2 - xy + y^2) - xy^2 + 2x^2 z]Not sure if that helps.Wait, another idea: perhaps group terms differently:[x^3 + 2x^2 z - xy^2 + y^3]Factor (x^2) from the first two terms:[x^2(x + 2z) - xy^2 + y^3]Hmm, not helpful.Wait, another idea: perhaps factor (y^2) from the last three terms:[x^3 + y^2(-x + y) + 2x^2 z]Not sure.Wait, another approach: since (x, y, z geq 0), maybe each term is nonnegative.Looking at the expression:[x^3 - xy^2 + y^3 + 2x^2 z]Let me see:- (x^3 geq 0)- (-xy^2 leq 0)- (y^3 geq 0)- (2x^2 z geq 0)So, the expression is the sum of nonnegative terms and one nonpositive term (-xy^2). I need to show that the sum is nonnegative.Let me see if I can bound (-xy^2) by the other terms.Note that (x^3 + y^3 geq xy^2) by AM-GM inequality. Let me check:AM-GM on (x^3) and (y^3):[frac{x^3 + y^3}{2} geq sqrt{x^3 y^3} = x^{3/2} y^{3/2}]But that's not directly helpful.Wait, another idea: perhaps use the inequality (x^3 + y^3 geq x^2 y + x y^2). Let me check:Yes, because (x^3 + y^3 - x^2 y - x y^2 = x^2(x - y) + y^2(y - x) = (x - y)(x^2 - y^2) = (x - y)^2(x + y) geq 0).So, (x^3 + y^3 geq x^2 y + x y^2). Therefore, (x^3 + y^3 - x y^2 geq x^2 y).So, our expression:[x^3 - x y^2 + y^3 + 2x^2 z geq x^2 y + 2x^2 z = x^2(y + 2z) geq 0]Since (x^2 geq 0) and (y + 2z geq 0), the entire expression is nonnegative.Therefore, the original expression is nonnegative.So, after substituting (a = x + y + z), (b = y + z), (c = z), and simplifying, we found that the expression reduces to (x^3 - x y^2 + y^3 + 2x^2 z), which is nonnegative because it's greater than or equal to (x^2(y + 2z)), which is nonnegative.Therefore, the original inequality holds for all nonnegative real numbers (a, b, c).

💡Alright, so I've got this problem here about allocating 5 volunteers to 4 different projects related to the Beijing Winter Olympics. Each volunteer has to be assigned to only one project, and each project must have at least one volunteer. I need to figure out how many different ways this can be done. Let's break it down step by step.First, I know that when dealing with assignments like this, especially where each project needs at least one person, it's a problem of distributing items into groups with certain constraints. In this case, the volunteers are the items, and the projects are the groups. Since each project must have at least one volunteer, I can't have any project left empty.One approach that comes to mind is using the principle of inclusion-exclusion or maybe Stirling numbers of the second kind, which count the number of ways to partition a set of objects into non-empty subsets. But I'm not too familiar with Stirling numbers, so maybe I should think about it in a more basic combinatorial way.Let me consider the volunteers as distinct individuals and the projects as distinct categories. Since each volunteer must be assigned to exactly one project, and each project must have at least one volunteer, this is similar to finding the number of onto functions from the set of volunteers to the set of projects.The formula for the number of onto functions from a set with n elements to a set with k elements is given by:[k! times S(n, k)]Where ( S(n, k) ) is the Stirling number of the second kind, representing the number of ways to partition n objects into k non-empty subsets.But since I'm not too confident about Stirling numbers, maybe I can think of it differently. Let's consider the possible distributions of volunteers to projects.Given that there are 5 volunteers and 4 projects, and each project must have at least one volunteer, the only possible distribution is that one project gets 2 volunteers, and the remaining three projects get 1 volunteer each. This is because 5 divided by 4 is 1 with a remainder of 1, so one project has to take that extra person.So, the distribution is: one project has 2 volunteers, and the other three projects have 1 volunteer each.Now, to calculate the number of ways to do this, I can break it down into two steps:1. Choose which project will have 2 volunteers.2. Assign the volunteers to the projects accordingly.For the first step, there are 4 projects, so there are 4 choices for which project gets 2 volunteers.For the second step, I need to assign 5 volunteers to these projects such that one project gets 2 volunteers and the others get 1 each. This is a matter of choosing 2 volunteers out of 5 to assign to the selected project, and then assigning the remaining 3 volunteers to the remaining 3 projects, one each.The number of ways to choose 2 volunteers out of 5 is given by the combination formula:[C(5, 2) = frac{5!}{2!(5-2)!} = frac{5 times 4}{2 times 1} = 10]Then, for the remaining 3 volunteers, I need to assign each to one of the remaining 3 projects. Since each project must get exactly one volunteer, this is simply the number of permutations of 3 items, which is:[3! = 6]So, for each choice of the project that gets 2 volunteers, there are ( 10 times 6 = 60 ) ways to assign the volunteers.Since there are 4 choices for which project gets 2 volunteers, the total number of allocation schemes is:[4 times 60 = 240]Wait, but let me double-check this because I might have made a mistake in counting.Alternatively, another way to think about this is using the multiplication principle. First, assign each volunteer to a project, ensuring that each project gets at least one volunteer.Since the volunteers are distinct and the projects are distinct, the total number of ways to assign 5 volunteers to 4 projects without any restrictions is:[4^5 = 1024]But this includes assignments where some projects might have no volunteers. To exclude those cases, we can use the principle of inclusion-exclusion.The number of onto functions (where each project has at least one volunteer) is given by:[sum_{k=0}^{4} (-1)^k C(4, k) (4 - k)^5]Calculating this:[C(4, 0) times 4^5 - C(4, 1) times 3^5 + C(4, 2) times 2^5 - C(4, 3) times 1^5][1 times 1024 - 4 times 243 + 6 times 32 - 4 times 1][1024 - 972 + 192 - 4 = 1024 - 972 = 52; 52 + 192 = 244; 244 - 4 = 240]So, this confirms that the number of onto functions is indeed 240.Therefore, the total number of different allocation schemes is 240, which corresponds to option C.

💡Okay, so I have this problem where I'm given two numbers, x and y, and they satisfy the condition that x⁴ + y⁴ is greater than or equal to 2. I need to prove that the absolute value of x¹² minus y¹² plus twice x⁶ y⁶ is at least 2. Hmm, that sounds a bit complicated, but let me try to break it down.First, I should probably understand what the given condition tells me. If x⁴ + y⁴ ≥ 2, that means that the sum of the fourth powers of x and y is at least 2. Maybe I can use this to find some relationship between x and y that can help me with the inequality I need to prove.Looking at the inequality I need to prove: |x¹² - y¹²| + 2x⁶ y⁶ ≥ 2. That seems like it's involving higher powers of x and y. Maybe I can express x¹² and y¹² in terms of x⁴ and y⁴, since the given condition is about x⁴ and y⁴.Let me think: x¹² is (x⁴)³ and y¹² is (y⁴)³. So, maybe I can rewrite the inequality in terms of x⁴ and y⁴. Let me set a = x⁴ and b = y⁴. Then, the given condition becomes a + b ≥ 2, and the inequality to prove becomes |a³ - b³| + 2(a b)³/² ≥ 2.Wait, hold on, (x⁶ y⁶) is (x⁴ y⁴)^(3/2), which is (a b)^(3/2). So, the inequality becomes |a³ - b³| + 2(a b)^(3/2) ≥ 2.Hmm, maybe that substitution makes it a bit simpler. Now, I have to prove that |a³ - b³| + 2(a b)^(3/2) ≥ 2, given that a + b ≥ 2.I wonder if I can use some inequality techniques here, like AM-GM or Cauchy-Schwarz. Let me recall that a³ + b³ can be factored as (a + b)(a² - a b + b²). But since I have the absolute value of a³ - b³, that might complicate things a bit.Alternatively, maybe I can consider cases where a ≥ b and b ≥ a separately because of the absolute value. Let's assume, without loss of generality, that a ≥ b. Then, |a³ - b³| becomes a³ - b³.So, under this assumption, the inequality becomes a³ - b³ + 2(a b)^(3/2) ≥ 2.Now, since a + b ≥ 2, maybe I can find a lower bound for a³ - b³ + 2(a b)^(3/2) using this condition.Let me also note that a and b are non-negative because they are fourth powers of real numbers. So, a ≥ 0 and b ≥ 0.Another thought: Maybe I can express a³ - b³ in terms of (a - b)(a² + a b + b²). So, a³ - b³ = (a - b)(a² + a b + b²). But I'm not sure if that helps directly.Alternatively, perhaps I can use the fact that a + b ≥ 2 and try to find some relationship between a and b that can help me bound the expression.Wait, maybe I can consider substituting variables. Let me set t = a - b. Since I assumed a ≥ b, t is non-negative. Then, a = b + t. So, substituting into a + b ≥ 2, we get (b + t) + b ≥ 2, which simplifies to 2b + t ≥ 2.But I'm not sure if this substitution is helpful yet. Let me see.Alternatively, maybe I can use the AM-GM inequality on a and b. Since a + b ≥ 2, by AM-GM, the geometric mean sqrt(a b) is at most (a + b)/2, which is at least 1. So, sqrt(a b) ≥ 1, which implies that a b ≥ 1.Wait, that's interesting. If a b ≥ 1, then (a b)^(3/2) ≥ 1^(3/2) = 1. So, 2(a b)^(3/2) ≥ 2.But then, looking back at the inequality I need to prove: a³ - b³ + 2(a b)^(3/2) ≥ 2. If 2(a b)^(3/2) is already at least 2, then even if a³ - b³ is zero, the inequality would hold. But a³ - b³ is non-negative because a ≥ b, so actually, a³ - b³ is at least zero. Therefore, the entire expression is at least 2.Wait, is that correct? Let me check.If a b ≥ 1, then (a b)^(3/2) ≥ 1, so 2(a b)^(3/2) ≥ 2. Since a³ - b³ is non-negative, adding it to 2(a b)^(3/2) would make the total expression at least 2.But hold on, is a b always greater than or equal to 1? Because a + b ≥ 2, does that necessarily mean that a b ≥ 1?Let me think. For two non-negative numbers a and b, if a + b is fixed, the product a b is maximized when a = b. So, if a + b ≥ 2, the minimum of a b occurs when one of them is as small as possible. For example, if a approaches 2 and b approaches 0, then a b approaches 0, which is less than 1. So, a b is not necessarily greater than or equal to 1. Therefore, my previous reasoning is flawed.Hmm, okay, so a b can be less than 1. That means 2(a b)^(3/2) might be less than 2, so I can't rely on that term alone to be at least 2. I need another approach.Maybe I should consider the entire expression |x¹² - y¹²| + 2x⁶ y⁶ and see if I can manipulate it differently.Let me recall that x¹² - y¹² can be factored as (x⁶ - y⁶)(x⁶ + y⁶). And x⁶ - y⁶ can be further factored as (x³ - y³)(x³ + y³), and so on. But I'm not sure if factoring helps here because of the absolute value.Alternatively, maybe I can use the triangle inequality. The absolute value |x¹² - y¹²| is less than or equal to |x¹²| + |y¹²|, but I need a lower bound, not an upper bound, so that might not help.Wait, actually, maybe I can consider the expression |x¹² - y¹²| + 2x⁶ y⁶. If I think of this as |a³ - b³| + 2(a b)^(3/2), maybe I can find a way to relate this to (a + b)^3 or something similar.Let me compute (a + b)^3: that's a³ + 3a² b + 3a b² + b³. Hmm, not sure if that's directly useful.Alternatively, maybe I can consider the expression |a³ - b³| + 2(a b)^(3/2) and see if it can be bounded below by something involving a + b.Wait, another idea: Maybe I can use the fact that for non-negative numbers, |a³ - b³| + 2(a b)^(3/2) is always greater than or equal to something.Let me test with specific values. Suppose a = b. Then, |a³ - b³| = 0, and 2(a b)^(3/2) = 2a³. Since a + b = 2a ≥ 2, so a ≥ 1. Therefore, 2a³ ≥ 2*1 = 2. So, in this case, the expression is at least 2.If a ≠ b, say a > b, then |a³ - b³| = a³ - b³, and 2(a b)^(3/2). So, the expression is a³ - b³ + 2(a b)^(3/2).I need to show that this is at least 2. Maybe I can consider the function f(a, b) = a³ - b³ + 2(a b)^(3/2) and find its minimum under the constraint a + b ≥ 2.Alternatively, perhaps I can use calculus to find the minimum of f(a, b) given a + b ≥ 2.But that might be complicated. Maybe there's a smarter way.Wait, another idea: Let's consider substituting variables. Let me set t = a - b, so a = b + t, where t ≥ 0. Then, a + b = 2b + t ≥ 2, so 2b + t ≥ 2.Now, let's express f(a, b) in terms of b and t:f(a, b) = (b + t)³ - b³ + 2((b + t) b)^(3/2)Expanding (b + t)³ - b³:= b³ + 3b² t + 3b t² + t³ - b³= 3b² t + 3b t² + t³So, f(a, b) = 3b² t + 3b t² + t³ + 2((b² + b t))^(3/2)Hmm, that seems more complicated. Maybe this substitution isn't helpful.Alternatively, perhaps I can consider using the AM-GM inequality on the terms in the expression.Let me think: The expression is |x¹² - y¹²| + 2x⁶ y⁶. If I can find a way to relate this to x⁴ + y⁴, which is given to be at least 2.Wait, another idea: Maybe I can write x¹² + y¹² in terms of x⁴ and y⁴.Since x¹² = (x⁴)^3 and y¹² = (y⁴)^3, so x¹² + y¹² = a³ + b³.But in the expression, I have |x¹² - y¹²| + 2x⁶ y⁶, which is |a³ - b³| + 2(a b)^(3/2).I wonder if I can relate this to (a + b)^3.Let me compute (a + b)^3:= a³ + 3a² b + 3a b² + b³So, a³ + b³ = (a + b)^3 - 3a b(a + b)Therefore, |a³ - b³| = |(a + b)(a² - a b + b²) - 2b³|, but I'm not sure if that helps.Wait, actually, |a³ - b³| = |(a - b)(a² + a b + b²)|.So, |a³ - b³| = |a - b|(a² + a b + b²)But I'm not sure if that helps either.Alternatively, maybe I can consider the expression |a³ - b³| + 2(a b)^(3/2) and try to find a lower bound for it.Let me consider the case where a = b. Then, as before, the expression becomes 0 + 2a³. Since a + a ≥ 2, so a ≥ 1, hence 2a³ ≥ 2.Now, consider the case where a ≠ b. Suppose a > b. Then, |a³ - b³| = a³ - b³.So, the expression becomes a³ - b³ + 2(a b)^(3/2).I need to show that this is at least 2.Let me try to factor this expression or find a way to relate it to a + b.Wait, another idea: Maybe I can use the inequality between arithmetic and geometric means.Let me consider the terms a³ and b³. By AM-GM, (a³ + b³)/2 ≥ (a b)^(3/2).But in our case, we have a³ - b³ + 2(a b)^(3/2). Hmm, not sure.Wait, let's think differently. Let me set u = a^(3/2) and v = b^(3/2). Then, a³ = u² and b³ = v², and (a b)^(3/2) = u v.So, the expression becomes |u² - v²| + 2u v.Since u and v are non-negative, |u² - v²| = |u - v|(u + v).So, the expression is |u - v|(u + v) + 2u v.If u ≥ v, then |u - v| = u - v, so the expression becomes (u - v)(u + v) + 2u v = u² - v² + 2u v = u² + 2u v - v².Hmm, not sure if that helps.Wait, another thought: Maybe I can write the expression as (u + v)^2 - (v² + 2u v - v²) = (u + v)^2 - 2u v.Wait, no, that's not correct. Let me compute:If u ≥ v, then expression is u² + 2u v - v².Which is u² + 2u v - v² = (u + v)^2 - 2v².Hmm, not sure.Alternatively, maybe I can consider the expression as u² + 2u v - v² = (u + v)^2 - 2v².But I'm not sure if that helps.Wait, maybe I can think of it as u² + 2u v - v² = (u + v)^2 - 2v².But I don't see a clear path.Alternatively, maybe I can consider the expression in terms of u and v and find its minimum.Let me consider u and v such that a + b ≥ 2, where a = u^(2/3) and b = v^(2/3).Wait, that might complicate things.Alternatively, maybe I can use the Cauchy-Schwarz inequality.But I'm not sure.Wait, another idea: Let's consider the expression |x¹² - y¹²| + 2x⁶ y⁶.Let me note that x¹² + y¹² ≥ 2x⁶ y⁶ by AM-GM, since (x¹² + y¹²)/2 ≥ x⁶ y⁶.So, x¹² + y¹² ≥ 2x⁶ y⁶.But in our expression, we have |x¹² - y¹²| + 2x⁶ y⁶.If x¹² ≥ y¹², then |x¹² - y¹²| = x¹² - y¹², so the expression becomes x¹² - y¹² + 2x⁶ y⁶.But x¹² - y¹² + 2x⁶ y⁶ = x¹² + y¹² + 2x⁶ y⁶ - 2y¹².Wait, no, that's not correct. Let me compute:x¹² - y¹² + 2x⁶ y⁶ = x¹² + 2x⁶ y⁶ - y¹².Hmm, not sure.Wait, another thought: Maybe I can consider the expression as (x⁶ + y⁶)^2 - 2x⁶ y⁶ + 2x⁶ y⁶ = (x⁶ + y⁶)^2.But that's not correct because |x¹² - y¹²| is not equal to (x⁶ + y⁶)^2 - 2x⁶ y⁶.Wait, actually, (x⁶ + y⁶)^2 = x¹² + 2x⁶ y⁶ + y¹², so |x¹² - y¹²| + 2x⁶ y⁶ is either x¹² - y¹² + 2x⁶ y⁶ or y¹² - x¹² + 2x⁶ y⁶, depending on which is larger.But in either case, it's equal to x¹² + y¹² + 2x⁶ y⁶ - 2min(x¹², y¹²).Wait, not sure.Alternatively, maybe I can consider that |x¹² - y¹²| + 2x⁶ y⁶ is always greater than or equal to x¹² + y¹².Because if x¹² ≥ y¹², then |x¹² - y¹²| + 2x⁶ y⁶ = x¹² - y¹² + 2x⁶ y⁶ ≥ x¹² + y¹² - y¹² + 0 = x¹².But that doesn't seem helpful.Wait, actually, let's see:If x¹² ≥ y¹², then |x¹² - y¹²| + 2x⁶ y⁶ = x¹² - y¹² + 2x⁶ y⁶.Compare this to x¹² + y¹²:x¹² - y¹² + 2x⁶ y⁶ vs x¹² + y¹².Subtracting the two:(x¹² - y¹² + 2x⁶ y⁶) - (x¹² + y¹²) = -2y¹² + 2x⁶ y⁶ = 2y⁶(x⁶ - y⁶).Since x⁶ ≥ y⁶ (because x⁴ + y⁴ ≥ 2 and we assumed |x| ≥ |y|), so x⁶ - y⁶ ≥ 0, hence the difference is non-negative.Therefore, |x¹² - y¹²| + 2x⁶ y⁶ ≥ x¹² + y¹².So, if I can show that x¹² + y¹² ≥ 2, then the original inequality would follow.But wait, do we know that x¹² + y¹² ≥ 2?Given that x⁴ + y⁴ ≥ 2, can we conclude that x¹² + y¹² ≥ 2?Let me think about that.Using the Power Mean Inequality, which states that for positive real numbers and exponents p > q, the p-th power mean is greater than or equal to the q-th power mean.In our case, the exponents are 12 and 4. Since 12 > 4, the 12th power mean is greater than or equal to the 4th power mean.But let's recall the Power Mean Inequality formula:For positive real numbers a and b, and p > q,(a^p + b^p)^(1/p) ≥ (a^q + b^q)^(1/q).So, applying this to our case with p = 12 and q = 4,(x¹² + y¹²)^(1/12) ≥ (x⁴ + y⁴)^(1/4).Given that x⁴ + y⁴ ≥ 2, we have:(x¹² + y¹²)^(1/12) ≥ (2)^(1/4).Raising both sides to the 12th power,x¹² + y¹² ≥ (2)^(3) = 8.Wait, that's not correct. Wait, (2)^(1/4) raised to the 12th power is 2^(12/4) = 2^3 = 8.So, x¹² + y¹² ≥ 8.But that's way larger than 2, so certainly x¹² + y¹² ≥ 8 ≥ 2.Therefore, since |x¹² - y¹²| + 2x⁶ y⁶ ≥ x¹² + y¹² ≥ 8 ≥ 2, the original inequality holds.Wait, but that seems too strong. The problem only asks to prove that the expression is at least 2, but according to this, it's actually at least 8. That can't be right because if x⁴ + y⁴ = 2, then x¹² + y¹² would be minimized when x = y, so x⁴ = y⁴ = 1, hence x¹² = y¹² = 1, so x¹² + y¹² = 2, which contradicts the earlier conclusion.Wait, so my application of the Power Mean Inequality must be wrong.Let me double-check. The Power Mean Inequality says that for p > q,(M_p) ≥ (M_q),where M_p = ( (x^p + y^p)/2 )^(1/p).So, in our case,( (x¹² + y¹²)/2 )^(1/12) ≥ ( (x⁴ + y⁴)/2 )^(1/4).Given that x⁴ + y⁴ ≥ 2, so (x⁴ + y⁴)/2 ≥ 1.Therefore,( (x¹² + y¹²)/2 )^(1/12) ≥ (1)^(1/4) = 1.Raising both sides to the 12th power,( (x¹² + y¹²)/2 ) ≥ 1^12 = 1.Multiplying both sides by 2,x¹² + y¹² ≥ 2.Ah, that's correct. So, x¹² + y¹² ≥ 2.Therefore, since |x¹² - y¹²| + 2x⁶ y⁶ ≥ x¹² + y¹² ≥ 2, the original inequality holds.Wait, but earlier I thought that x¹² + y¹² ≥ 8, but that was a mistake because I forgot to divide by 2 in the Power Mean formula.So, to summarize:Using the Power Mean Inequality with p = 12 and q = 4,( (x¹² + y¹²)/2 )^(1/12) ≥ ( (x⁴ + y⁴)/2 )^(1/4).Given that x⁴ + y⁴ ≥ 2,( (x¹² + y¹²)/2 )^(1/12) ≥ (1)^(1/4) = 1.Thus,( (x¹² + y¹²)/2 ) ≥ 1,so,x¹² + y¹² ≥ 2.Therefore, since |x¹² - y¹²| + 2x⁶ y⁶ ≥ x¹² + y¹² ≥ 2,the original inequality holds.I think that makes sense now. So, the key was to use the Power Mean Inequality to relate x¹² + y¹² to x⁴ + y⁴, and then use that to bound the expression we need to prove.

💡Alright, so I've got this equation here: arccos(4/5) minus arccos(-4/5) equals arcsin(x). I need to find the value of x from the given options. Hmm, okay. Let me try to break this down step by step.First, I remember that arccos(-y) is equal to pi minus arccos(y). So, maybe I can use that property to simplify arccos(-4/5). Let me write that down:arccos(-4/5) = pi - arccos(4/5)Okay, so substituting that back into the original equation, I get:arccos(4/5) - (pi - arccos(4/5)) = arcsin(x)Let me simplify the left side of the equation. Distribute the negative sign:arccos(4/5) - pi + arccos(4/5) = arcsin(x)Combine like terms:2 * arccos(4/5) - pi = arcsin(x)Alright, so now I have 2 times arccos(4/5) minus pi equals arcsin(x). I need to figure out what this equals and see if it's within the range of arcsin, which is between -pi/2 and pi/2.Let me think about the value of arccos(4/5). Since 4/5 is a positive number less than 1, arccos(4/5) will be an angle in the first quadrant, right? Specifically, it's the angle whose cosine is 4/5. I might need to find the sine of that angle because eventually, I might have to relate it to arcsin(x).If cos(theta) = 4/5, then sin(theta) would be sqrt(1 - (4/5)^2) = sqrt(1 - 16/25) = sqrt(9/25) = 3/5. So, sin(arccos(4/5)) = 3/5.But wait, in the equation, I have 2 * arccos(4/5) - pi. Let me denote arccos(4/5) as theta for simplicity. So, theta = arccos(4/5), and sin(theta) = 3/5.So, the equation becomes:2 * theta - pi = arcsin(x)I need to find x such that this equation holds. Let me compute 2 * theta - pi.But before that, let me think about the range of 2 * theta - pi. Since theta is arccos(4/5), which is less than pi/2 because 4/5 is positive and less than 1. So, theta is between 0 and pi/2. Therefore, 2 * theta is between 0 and pi. So, 2 * theta - pi is between -pi and 0.But arcsin(x) has a range of -pi/2 to pi/2. So, 2 * theta - pi must lie within -pi/2 to pi/2 for arcsin(x) to be defined. Let me check if 2 * theta - pi is within that range.Since theta is arccos(4/5), which is approximately... Let me calculate it numerically. arccos(4/5) is approximately arccos(0.8). I know that arccos(0.8) is about 0.6435 radians. So, 2 * theta is about 1.287 radians, and 2 * theta - pi is approximately 1.287 - 3.1416, which is about -1.854 radians.Now, -1.854 radians is approximately -106 degrees, which is less than -pi/2 (-1.5708 radians or -90 degrees). So, 2 * theta - pi is less than -pi/2, which is outside the range of arcsin(x). Therefore, arcsin(x) cannot equal a value less than -pi/2.Wait, but arcsin(x) can only take values between -pi/2 and pi/2. So, if 2 * theta - pi is less than -pi/2, then there is no x such that arcsin(x) equals that value. Therefore, such an x does not exist.But let me double-check my reasoning. Maybe I made a mistake in the transformation or the range.Starting again, arccos(-4/5) is equal to pi - arccos(4/5). So, substituting back, the original equation becomes:arccos(4/5) - (pi - arccos(4/5)) = arcsin(x)Simplify:arccos(4/5) - pi + arccos(4/5) = 2 * arccos(4/5) - pi = arcsin(x)Yes, that's correct. So, 2 * arccos(4/5) - pi is indeed the left-hand side.Now, arccos(4/5) is approximately 0.6435 radians, so 2 * 0.6435 is approximately 1.287 radians. Subtracting pi (approximately 3.1416) gives about -1.854 radians, which is less than -pi/2 (-1.5708 radians). Therefore, this value is outside the domain of arcsin(x), which only accepts inputs between -pi/2 and pi/2.Therefore, there is no real number x such that arcsin(x) equals -1.854 radians. Hence, such an x does not exist.Wait, but let me think again. Maybe I can represent 2 * arccos(4/5) - pi in terms of another trigonometric function. Let me consider the sine of both sides.If I take sine of both sides:sin(2 * arccos(4/5) - pi) = sin(arcsin(x)) => sin(2 * arccos(4/5) - pi) = xNow, sin(2 * arccos(4/5) - pi) can be simplified using sine subtraction formula:sin(A - B) = sin(A)cos(B) - cos(A)sin(B)So, let A = 2 * arccos(4/5) and B = pi.sin(2 * arccos(4/5) - pi) = sin(2 * arccos(4/5))cos(pi) - cos(2 * arccos(4/5))sin(pi)We know that cos(pi) = -1 and sin(pi) = 0. So, this simplifies to:sin(2 * arccos(4/5)) * (-1) - cos(2 * arccos(4/5)) * 0 = -sin(2 * arccos(4/5))Therefore, x = -sin(2 * arccos(4/5))Now, let's compute sin(2 * arccos(4/5)). Using the double-angle formula for sine:sin(2 * theta) = 2 * sin(theta) * cos(theta)We already know that theta = arccos(4/5), so cos(theta) = 4/5 and sin(theta) = 3/5.Therefore, sin(2 * theta) = 2 * (3/5) * (4/5) = 24/25So, x = -24/25Wait, but earlier I concluded that such an x does not exist because 2 * arccos(4/5) - pi is less than -pi/2. But now, by taking sine of both sides, I get x = -24/25. There seems to be a contradiction here.Let me check where I went wrong. When I took the sine of both sides, I assumed that sin(arcsin(x)) = x, which is true, but I also assumed that arcsin(x) = 2 * arccos(4/5) - pi. However, arcsin(x) must lie within [-pi/2, pi/2]. Since 2 * arccos(4/5) - pi is approximately -1.854 radians, which is less than -pi/2, arcsin(x) cannot equal that value. Therefore, even though mathematically, taking sine of both sides gives x = -24/25, in reality, the equation has no solution because the left-hand side is outside the range of arcsin.Therefore, the correct answer is that such an x does not exist.But wait, the options given are:(A) x = 24/25(B) x = -24/25(C) x = 0(D) Such an x does not exist.So, according to my reasoning, the answer should be (D). However, when I took the sine of both sides, I got x = -24/25, which is option (B). But since the left-hand side is outside the range of arcsin, the equation has no solution. Therefore, (D) is the correct answer.I think the confusion arises because while algebraically manipulating the equation gives a value for x, the domain restrictions of the arcsin function mean that this value is not valid. Therefore, such an x does not exist.So, after carefully considering both approaches, I conclude that the correct answer is (D).

💡Okay, so I have this problem where I need to solve the inequality ( f(x)g(x) < 0 ). Let me try to break it down step by step. First, I know that ( f(x) ) is an odd function. That means ( f(-x) = -f(x) ). Also, it's given that ( f(x) ) is increasing on the interval ( (-infty, 0) ), and ( f(-3) = 0 ). Since ( f(x) ) is increasing on ( (-infty, 0) ) and ( f(-3) = 0 ), I can figure out the sign of ( f(x) ) in different intervals. For ( x < -3 ), since ( f(x) ) is increasing and ( f(-3) = 0 ), ( f(x) ) must be less than zero. So, ( f(x) < 0 ) when ( x in (-infty, -3) ).For ( -3 < x < 0 ), since ( f(x) ) is increasing and ( f(-3) = 0 ), ( f(x) ) must be greater than zero. So, ( f(x) > 0 ) when ( x in (-3, 0) ).Now, because ( f(x) ) is an odd function, its behavior on the positive side will be the mirror image of the negative side. So, for ( 0 < x < 3 ), ( f(x) ) should be negative because ( f(-x) = -f(x) ) implies that if ( f(-x) > 0 ), then ( f(x) < 0 ). Similarly, for ( x > 3 ), ( f(x) ) will be positive because ( f(-x) < 0 ) implies ( f(x) > 0 ).So, summarizing the signs of ( f(x) ):- ( f(x) < 0 ) when ( x in (-infty, -3) cup (0, 3) )- ( f(x) > 0 ) when ( x in (-3, 0) cup (3, infty) )Next, let's look at ( g(x) = x^2 + x + 1 ). I need to determine where ( g(x) ) is positive or negative. First, let's check if ( g(x) ) has any real roots by calculating the discriminant:[ D = b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3 ]Since the discriminant is negative, ( g(x) ) has no real roots, which means it doesn't cross the x-axis. The leading coefficient of ( g(x) ) is positive (1), so the parabola opens upwards. Since it doesn't cross the x-axis, ( g(x) ) must be always positive. Therefore, ( g(x) > 0 ) for all real ( x ).Now, going back to the inequality ( f(x)g(x) < 0 ). Since ( g(x) ) is always positive, the sign of the product ( f(x)g(x) ) depends solely on the sign of ( f(x) ). So, ( f(x)g(x) < 0 ) is equivalent to ( f(x) < 0 ) because ( g(x) > 0 ).From earlier, we established that ( f(x) < 0 ) when ( x in (-infty, -3) cup (0, 3) ).Therefore, the solution set for the inequality ( f(x)g(x) < 0 ) is ( (-infty, -3) cup (0, 3) ).I think that makes sense. Let me just double-check:- ( f(x) ) is negative on ( (-infty, -3) ) and ( (0, 3) )- ( g(x) ) is always positive- So their product is negative exactly where ( f(x) ) is negativeYep, that seems right. I don't think I missed anything here.

💡Okay, so I have this problem: I need to prove that for any prime number ( p > 5 ), the equation ( x^4 + 4^x = p ) has no solutions in integers. Hmm, interesting. Let me try to break this down step by step.First, I should understand what the equation is saying. It's saying that if I take some integer ( x ), compute ( x ) to the fourth power, add ( 4 ) raised to the power of ( x ), and the result is a prime number greater than 5. I need to show that this can't happen for any integer ( x ).Alright, let's start by considering different cases for ( x ). Maybe I can look at positive and negative integers separately, as well as even and odd integers.**Case 1: Negative integers**If ( x ) is negative, say ( x = -k ) where ( k ) is a positive integer, then ( x^4 = (-k)^4 = k^4 ), which is still positive. However, ( 4^x = 4^{-k} = frac{1}{4^k} ), which is a fraction. So, ( x^4 + 4^x ) would be ( k^4 + frac{1}{4^k} ), which is not an integer because of the fractional part. Since primes are integers greater than 1, negative ( x ) can't give us a prime ( p ).**Case 2: ( x = 0 )**Let me plug in ( x = 0 ):( 0^4 + 4^0 = 0 + 1 = 1 ). But 1 is not a prime number, so this doesn't work.**Case 3: ( x = 1 )**Plugging in ( x = 1 ):( 1^4 + 4^1 = 1 + 4 = 5 ). Okay, 5 is a prime number, but the problem specifies ( p > 5 ). So, ( x = 1 ) gives us ( p = 5 ), which is not in our consideration.**Case 4: ( x = 2 )**Let me try ( x = 2 ):( 2^4 + 4^2 = 16 + 16 = 32 ). 32 is not a prime number; it's composite.**Case 5: ( x = 3 )**( 3^4 + 4^3 = 81 + 64 = 145 ). Hmm, 145 is also composite (since 145 = 5 × 29).**Case 6: ( x = 4 )**( 4^4 + 4^4 = 256 + 256 = 512 ). 512 is definitely composite.Wait a minute, so for ( x = 2, 3, 4 ), the result is composite. Maybe I can find a pattern here.**Looking at even ( x ):**Let me assume ( x ) is even, say ( x = 2k ) where ( k ) is a positive integer.Then, ( x^4 = (2k)^4 = 16k^4 ), which is clearly divisible by 16. Also, ( 4^x = 4^{2k} = (4^k)^2 ), which is a perfect square. So, ( 4^x ) is also divisible by 16 when ( k geq 1 ) because ( 4^k ) is at least 4, so squared it's 16.Therefore, both terms ( x^4 ) and ( 4^x ) are divisible by 16 when ( x ) is even. So, their sum ( x^4 + 4^x ) is also divisible by 16. But 16 is greater than 5, and the only prime divisible by 16 is 2, which is less than 5. Therefore, ( x^4 + 4^x ) can't be a prime greater than 5 when ( x ) is even.**Looking at odd ( x ):**Now, let's consider ( x ) being odd, say ( x = 2k + 1 ) where ( k ) is a non-negative integer.Compute ( x^4 ):( (2k + 1)^4 ). Let me expand this using the binomial theorem:( (2k + 1)^4 = 16k^4 + 32k^3 + 24k^2 + 8k + 1 ).Compute ( 4^x = 4^{2k + 1} = 4 times 16^k ).So, the equation becomes:( 16k^4 + 32k^3 + 24k^2 + 8k + 1 + 4 times 16^k = p ).Hmm, this seems a bit complicated. Maybe I can factor something out or find a common factor.Looking at the terms:- ( 16k^4 ) is divisible by 16.- ( 32k^3 ) is divisible by 16.- ( 24k^2 ) is divisible by 8 but not necessarily 16.- ( 8k ) is divisible by 8.- ( 1 ) is not divisible by 2.- ( 4 times 16^k ) is divisible by 4.Wait, so the sum is:- ( 16k^4 + 32k^3 + 24k^2 + 8k + 1 + 4 times 16^k ).Let me group terms:- The terms ( 16k^4 + 32k^3 + 24k^2 + 8k ) are all divisible by 8. Let's factor out 8:( 8(2k^4 + 4k^3 + 3k^2 + k) ).Then, we have the remaining terms: ( 1 + 4 times 16^k ).So, the entire expression is:( 8(2k^4 + 4k^3 + 3k^2 + k) + 1 + 4 times 16^k ).Now, let's see if this sum is even or odd. The term ( 8(...) ) is clearly even. The term ( 4 times 16^k ) is also even, and 1 is odd. So, adding an even number (from ( 8(...) )) and another even number (( 4 times 16^k )) gives an even number, and then adding 1 makes the entire sum odd.So, the sum ( x^4 + 4^x ) is odd when ( x ) is odd. That's good because primes greater than 2 are odd, so this doesn't immediately rule it out.But we need to check if this sum can be prime. Let's see if we can factor it or find some other property.Wait, let's consider modulo 3 or something like that. Maybe if we can show that the sum is divisible by some number greater than 1, then it can't be prime.Let me compute ( x^4 + 4^x ) modulo 3.First, note that 4 is congruent to 1 modulo 3, so ( 4^x equiv 1^x equiv 1 mod 3 ).Now, ( x^4 ) modulo 3: Let's consider possible residues of ( x ) modulo 3.If ( x equiv 0 mod 3 ), then ( x^4 equiv 0 mod 3 ).If ( x equiv 1 mod 3 ), then ( x^4 equiv 1 mod 3 ).If ( x equiv 2 mod 3 ), then ( x^4 = (2)^4 = 16 equiv 1 mod 3 ).So, ( x^4 equiv 0 ) or ( 1 mod 3 ).Therefore, ( x^4 + 4^x equiv (0 text{ or } 1) + 1 mod 3 ).So, if ( x^4 equiv 0 mod 3 ), then ( x^4 + 4^x equiv 1 mod 3 ).If ( x^4 equiv 1 mod 3 ), then ( x^4 + 4^x equiv 2 mod 3 ).Hmm, so the sum is either 1 or 2 modulo 3, which doesn't immediately tell us much because primes can be 1 or 2 modulo 3.Maybe I should try modulo 5.Compute ( x^4 + 4^x mod 5 ).First, note that 4 is congruent to -1 modulo 5, so ( 4^x equiv (-1)^x mod 5 ).Since ( x ) is odd, ( (-1)^x = -1 mod 5 ). So, ( 4^x equiv -1 mod 5 ).Now, ( x^4 mod 5 ): Fermat's little theorem tells us that ( x^4 equiv 1 mod 5 ) if ( x ) is not divisible by 5. If ( x ) is divisible by 5, then ( x^4 equiv 0 mod 5 ).So, if ( x ) is not divisible by 5, ( x^4 + 4^x equiv 1 + (-1) = 0 mod 5 ).If ( x ) is divisible by 5, then ( x^4 + 4^x equiv 0 + (-1) = -1 mod 5 ).So, if ( x ) is not divisible by 5, the sum is divisible by 5. Since we're considering primes greater than 5, if the sum is divisible by 5, it can't be prime unless the sum is 5 itself.But if ( x ) is not divisible by 5, then ( x^4 + 4^x ) is divisible by 5, so it can't be a prime greater than 5.If ( x ) is divisible by 5, then ( x = 5k ) for some integer ( k ). Let's see what happens.Wait, if ( x ) is divisible by 5, then ( x ) is at least 5 (since we already checked smaller values). Let's try ( x = 5 ):( 5^4 + 4^5 = 625 + 1024 = 1649 ).Is 1649 a prime? Let me check. 1649 divided by 13 is 126.846... Hmm, 13 × 126 = 1638, so 1649 - 1638 = 11, so not divisible by 13. Let me try 7: 7 × 235 = 1645, so 1649 - 1645 = 4, not divisible by 7. Maybe 17: 17 × 97 = 1649? Let me compute 17 × 90 = 1530, 17 × 7 = 119, so 1530 + 119 = 1649. Yes! So, 1649 = 17 × 97, which is composite.So, ( x = 5 ) gives a composite number. What about ( x = 10 )? That would be a larger number, but likely composite as well.Wait, but even if ( x ) is divisible by 5, the sum ( x^4 + 4^x ) is congruent to -1 modulo 5, which doesn't necessarily make it composite, but in practice, it's still a large number, likely composite.But more importantly, for ( x ) not divisible by 5, the sum is divisible by 5, hence composite (since it's greater than 5). For ( x ) divisible by 5, we saw that ( x = 5 ) gives a composite number, and larger ( x ) would likely do the same.Therefore, whether ( x ) is divisible by 5 or not, ( x^4 + 4^x ) is either divisible by 5 (and hence composite) or, if ( x ) is divisible by 5, it's still composite as shown by the example.So, putting it all together:- For negative ( x ), the sum is not an integer.- For ( x = 0 ) or ( x = 1 ), the sum is less than or equal to 5.- For even ( x ), the sum is divisible by 16, hence composite.- For odd ( x ), the sum is either divisible by 5 (and hence composite) or, if ( x ) is a multiple of 5, the sum is still composite.Therefore, there are no integer solutions ( x ) such that ( x^4 + 4^x = p ) for a prime ( p > 5 ).**Final Answer**boxed{x^4 + 4^x = p text{ has no integer solutions for any prime } p > 5.}