💡To find the largest possible value of ( ab + bc + cd + da ) where ( a, b, c, d ) are 2, 3, 5, and 6 in some order, we can start by expressing the given expression in a different form.First, notice that:[ab + bc + cd + da = b(a + c) + d(a + c) = (a + c)(b + d)]This transformation simplifies the problem to finding the maximum value of the product ( (a + c)(b + d) ).Given the numbers 2, 3, 5, and 6, their total sum is:[a + b + c + d = 2 + 3 + 5 + 6 = 16]We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to find the maximum product. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Applying this to ( (a + c) ) and ( (b + d) ):[frac{(a + c) + (b + d)}{2} geq sqrt{(a + c)(b + d)}][frac{16}{2} geq sqrt{(a + c)(b + d)}][8 geq sqrt{(a + c)(b + d)}][64 geq (a + c)(b + d)]Thus, the maximum possible value of ( (a + c)(b + d) ) is 64. However, we need to check if this value is achievable with the given numbers.Testing different assignments:- Let ( a = 2 ), ( b = 6 ), ( c = 5 ), ( d = 3 ): [ (a + c) = 2 + 5 = 7 ] [ (b + d) = 6 + 3 = 9 ] [ (a + c)(b + d) = 7 times 9 = 63 ]- Another assignment: ( a = 2 ), ( b = 3 ), ( c = 5 ), ( d = 6 ): [ (a + c) = 2 + 5 = 7 ] [ (b + d) = 3 + 6 = 9 ] [ (a + c)(b + d) = 7 times 9 = 63 ]Both assignments yield a product of 63, which is the highest achievable value with the given numbers.

💡First, I need to determine how much paint is used on the first day. Since one quarter of the paint is used, I calculate:[frac{1}{4} times 2 = frac{1}{2} text{ gallons}]Subtracting this from the initial amount gives the remaining paint after the first day:[2 - frac{1}{2} = frac{3}{2} text{ gallons}]Next, on the second day, one third of the remaining paint is used. I calculate the amount used on the second day as:[frac{1}{3} times frac{3}{2} = frac{1}{2} text{ gallons}]Subtracting this from the remaining paint after the first day gives the remaining paint after the second day:[frac{3}{2} - frac{1}{2} = 1 text{ gallon}]Finally, to find the fraction of the original amount of paint that is left, I compare the remaining paint to the original 2 gallons:[frac{1}{2} text{ of the original amount}]

💡Alright, so I have this problem here where I need to figure out which function among the options A to D has a different domain and range compared to the others. Let me take it step by step because I want to make sure I understand each function properly.First, let's list out the functions:A: ( y = x^{frac{1}{2}} ) B: ( y = x^{-1} ) C: ( y = x^{frac{1}{3}} ) D: ( y = x^{2} )I need to find the domain and range for each of these functions and then see which one stands out as different.Starting with option A: ( y = x^{frac{1}{2}} ). Hmm, this is the square root function. I remember that for square roots, the expression inside the root must be non-negative because you can't take the square root of a negative number in the set of real numbers. So, the domain here should be all real numbers ( x ) such that ( x geq 0 ). In interval notation, that's ( [0, +infty) ). Now, for the range. Since the square root function outputs only non-negative values, the smallest value ( y ) can be is 0, and it can go up to infinity. So, the range is also ( [0, +infty) ). So, both the domain and range for A are ( [0, +infty) ).Moving on to option B: ( y = x^{-1} ). That's the same as ( y = frac{1}{x} ). I recall that for this function, ( x ) cannot be zero because division by zero is undefined. So, the domain is all real numbers except zero, which I can write as ( (-infty, 0) cup (0, +infty) ).What about the range? Well, ( y ) can also never be zero because ( frac{1}{x} ) will never equal zero for any real ( x ). So, the range is also all real numbers except zero, ( (-infty, 0) cup (0, +infty) ). So, both the domain and range for B are ( (-infty, 0) cup (0, +infty) ).Next is option C: ( y = x^{frac{1}{3}} ). This is the cube root function. Unlike the square root, the cube root can handle negative numbers because the cube of a negative number is negative. So, the domain here is all real numbers, ( (-infty, +infty) ).For the range, since the cube root function can output any real number as well, the range is also all real numbers, ( (-infty, +infty) ). So, both the domain and range for C are ( (-infty, +infty) ).Finally, option D: ( y = x^{2} ). This is a quadratic function, specifically a parabola opening upwards. The domain for any polynomial function is all real numbers, so that's ( (-infty, +infty) ).Now, the range. Since squaring any real number gives a non-negative result, the smallest value ( y ) can be is 0, and it goes up to infinity. So, the range is ( [0, +infty) ). Let me summarize what I have:- **A**: Domain ( [0, +infty) ), Range ( [0, +infty) )- **B**: Domain ( (-infty, 0) cup (0, +infty) ), Range ( (-infty, 0) cup (0, +infty) )- **C**: Domain ( (-infty, +infty) ), Range ( (-infty, +infty) )- **D**: Domain ( (-infty, +infty) ), Range ( [0, +infty) )Looking at these, I need to find the function with a different domain and range. Let's compare them:- A has both domain and range as ( [0, +infty) )- B has both domain and range as ( (-infty, 0) cup (0, +infty) )- C has both domain and range as ( (-infty, +infty) )- D has domain ( (-infty, +infty) ) but range ( [0, +infty) )So, A, B, and C all have their domain and range matching each other in a way. For example, A and D both have domains that are not all real numbers, but D's range is different from A's. Wait, actually, A's domain is ( [0, +infty) ) and D's domain is all real numbers. So, D's domain is different from A, B, and C.But looking at the ranges, A has ( [0, +infty) ), B has ( (-infty, 0) cup (0, +infty) ), C has ( (-infty, +infty) ), and D also has ( [0, +infty) ). So, A and D share the same range, but different domains.Wait, so A has domain ( [0, +infty) ) and range ( [0, +infty) ). D has domain ( (-infty, +infty) ) and range ( [0, +infty) ). So, D's domain is different from A's, but its range is the same as A's.But B and C have different domains and ranges from A and D. So, is D the only one with a different domain and range compared to others? Or is it B?Wait, let's see:- A: Domain and range both ( [0, +infty) )- B: Domain and range both ( (-infty, 0) cup (0, +infty) )- C: Domain and range both ( (-infty, +infty) )- D: Domain ( (-infty, +infty) ), Range ( [0, +infty) )So, A, B, and C have their domain and range matching each other in some way—each function's domain and range are the same. However, D has a different range compared to its domain. So, D is the one where the domain and range are different from each other, whereas for A, B, and C, the domain and range are the same.Wait, but the question is asking for the function with a different domain and range compared to the others. So, are we comparing the domain and range of each function to each other, or are we comparing each function's domain and range to the others'?I think the question is asking which function has a different domain and range compared to the others. So, looking at all four functions, which one has a unique combination of domain and range.Looking at A: Domain and range both ( [0, +infty) )B: Domain and range both ( (-infty, 0) cup (0, +infty) )C: Domain and range both ( (-infty, +infty) )D: Domain ( (-infty, +infty) ), Range ( [0, +infty) )So, A, B, and C have their domain and range matching each other in the sense that each function's domain and range are the same. However, D has a domain that's all real numbers but a range that's only non-negative real numbers. So, D is the only function where the domain and range are not the same.Therefore, D is the function with a different domain and range compared to the others.But wait, let me double-check. The question says "the one with a different domain and range." So, is it that D's domain is different from the others, or its range is different?Looking back:- A's domain is ( [0, +infty) ), which is different from B, C, D.- B's domain is ( (-infty, 0) cup (0, +infty) ), different from A, C, D.- C's domain is ( (-infty, +infty) ), same as D.- D's domain is ( (-infty, +infty) ), same as C.So, A and B have unique domains, but C and D share the same domain.Similarly, for ranges:- A's range is ( [0, +infty) ), same as D.- B's range is ( (-infty, 0) cup (0, +infty) ), unique.- C's range is ( (-infty, +infty) ), unique.- D's range is ( [0, +infty) ), same as A.So, in terms of both domain and range:- A: Domain ( [0, +infty) ), Range ( [0, +infty) )- B: Domain ( (-infty, 0) cup (0, +infty) ), Range ( (-infty, 0) cup (0, +infty) )- C: Domain ( (-infty, +infty) ), Range ( (-infty, +infty) )- D: Domain ( (-infty, +infty) ), Range ( [0, +infty) )So, A, B, and C have their domain equal to their range, but D does not. So, D is the only function where the domain and range are different from each other, making it the one with a different domain and range compared to the others.Therefore, the answer should be D.**Final Answer**boxed{D}

💡Alright, so I've got this problem here about Ivan, Peter, and Sidor eating candies. Their last names are Ivanov, Petrov, and Sidorov. The problem gives me some clues:1. Ivanov ate 2 candies less than Ivan.2. Petrov ate 2 candies less than Peter.3. Peter ate the most.And I need to figure out who has which last name. Okay, let's break this down step by step.First, let's list out the people and their possible last names:- Ivan- Peter- SidorAnd the last names:- Ivanov- Petrov- SidorovSo, each person has one of these last names, and no two people share the same last name.Now, let's look at the clues:1. Ivanov ate 2 candies less than Ivan.2. Petrov ate 2 candies less than Peter.3. Peter ate the most.Alright, clue number 3 says Peter ate the most. So, Peter is the one who consumed the highest number of candies among the three.Clue number 1 says Ivanov ate 2 candies less than Ivan. So, Ivanov is someone who ate fewer candies than Ivan. Similarly, clue number 2 says Petrov ate 2 candies less than Peter. So, Petrov is someone who ate fewer candies than Peter.Since Peter ate the most, and Petrov ate 2 candies less than Peter, that would mean Petrov is not Peter. So, Petrov must be either Ivan or Sidor.Similarly, Ivanov ate 2 candies less than Ivan, so Ivanov is not Ivan. Therefore, Ivanov must be either Peter or Sidor.But wait, Peter is the one who ate the most, and Ivanov ate 2 candies less than Ivan. So, if Ivanov is Peter, that would mean Peter ate 2 candies less than Ivan, but we know Peter ate the most. That doesn't make sense because if Peter ate the most, he can't have eaten less than Ivan. So, Ivanov can't be Peter. Therefore, Ivanov must be Sidor.So, Sidor is Ivanov.Now, since Ivanov is Sidor, and Petrov is either Ivan or Sidor, but Sidor is already Ivanov, Petrov must be Ivan.So, Ivan is Petrov.That leaves Peter to be Sidorov.Let me recap:- Sidor is Ivanov.- Ivan is Petrov.- Peter is Sidorov.Does this make sense?Let's check the clues:1. Ivanov (Sidor) ate 2 candies less than Ivan (Petrov). So, Sidor ate 2 less than Petrov. That works.2. Petrov (Ivan) ate 2 candies less than Peter (Sidorov). So, Ivan ate 2 less than Peter. That also works.3. Peter (Sidorov) ate the most. Yes, that's consistent.So, it seems to fit.But wait, let me double-check to make sure I didn't make any mistakes.If Sidor is Ivanov, then Ivanov is Sidor. Ivanov ate 2 less than Ivan. So, Sidor ate 2 less than Ivan. That means Ivan ate more than Sidor.Petrov ate 2 less than Peter. So, Petrov is Ivan, and Ivan ate 2 less than Peter. So, Peter ate more than Ivan.And Peter ate the most, which is consistent.Yes, everything checks out.So, the final assignments are:- Ivan: Petrov- Peter: Sidorov- Sidor: IvanovI think that's the correct solution.

💡First, I identify the principal amount as ( P ) and the simple interest rate as ( r ).Given that the amount after 2 years is Rs. 720, I use the simple interest formula:[720 = P + (P times r times 2)]This simplifies to:[720 = P(1 + 2r)]Similarly, for the amount after a total of 7 years being Rs. 1020:[1020 = P + (P times r times 7)]Which simplifies to:[1020 = P(1 + 7r)]I now have two equations:1. ( 720 = P(1 + 2r) )2. ( 1020 = P(1 + 7r) )Subtracting the first equation from the second to eliminate ( P ):[1020 - 720 = P(1 + 7r) - P(1 + 2r)][300 = P(5r)][r = frac{300}{5P}]Substituting ( r ) back into the first equation:[720 = Pleft(1 + 2 times frac{300}{5P}right)][720 = Pleft(1 + frac{600}{5P}right)][720 = P + 120][P = 720 - 120][P = 600]Therefore, the principal amount is Rs. 600.

💡Okay, so I need to prove that for positive real numbers ( a, b, c ), the expression ( frac{a}{b+c} + frac{b}{c+a} + frac{c}{a+b} ) is greater than or equal to ( frac{3}{2} ). Hmm, I remember hearing about inequalities like this before, maybe something to do with the Nesbitt's inequality? I think that's the one. But I'm not entirely sure how to approach it. Let me think step by step.First, I should probably recall what Nesbitt's inequality actually states. If I remember correctly, it says that for positive real numbers ( a, b, c ), the following holds:[frac{a}{b + c} + frac{b}{c + a} + frac{c}{a + b} geq frac{3}{2}]Yes, that seems right. So, I need to prove this inequality. I wonder what methods are commonly used for such proofs. Maybe I can try using the AM-GM inequality or Cauchy-Schwarz. Let me explore these options.Starting with the AM-GM inequality. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. But I'm not sure how directly applicable that is here. Maybe I can manipulate the given expression to fit the form required for AM-GM.Alternatively, the Cauchy-Schwarz inequality is often useful in these situations. The Cauchy-Schwarz inequality in its basic form is:[left( sum_{i=1}^n x_i y_i right)^2 leq left( sum_{i=1}^n x_i^2 right) left( sum_{i=1}^n y_i^2 right)]But I'm not sure how to apply this directly to the given expression. Maybe I need to consider some form of substitution or rearrangement.Wait, another thought: perhaps I can use the concept of homogenization. Since all terms in the inequality are homogeneous of degree 0 (meaning if I scale ( a, b, c ) by a common factor, the inequality remains unchanged), I can assume without loss of generality that ( a + b + c = 1 ). That might simplify things a bit.So, let's set ( a + b + c = 1 ). Then, the denominators ( b + c ), ( c + a ), and ( a + b ) become ( 1 - a ), ( 1 - b ), and ( 1 - c ) respectively. So, the expression becomes:[frac{a}{1 - a} + frac{b}{1 - b} + frac{c}{1 - c}]Hmm, not sure if that helps directly. Maybe I can consider another approach. Let me think about the method of substitution or maybe even the Rearrangement Inequality.Wait, another idea: perhaps I can use the fact that ( frac{a}{b + c} ) can be rewritten as ( frac{a^2}{ab + ac} ). Then, by the Cauchy-Schwarz inequality, we have:[sum frac{a^2}{ab + ac} geq frac{(a + b + c)^2}{2(ab + bc + ca)}]Yes, that seems promising. Let me write that down more formally.Using the Cauchy-Schwarz inequality in the form:[left( sum frac{a^2}{x} right) geq frac{(a + b + c)^2}{sum x}]where ( x = ab + ac ), ( y = bc + ba ), and ( z = ca + cb ). Wait, no, actually, in this case, each denominator is ( ab + ac ), ( bc + ba ), and ( ca + cb ). So, the sum of the denominators would be ( 2(ab + bc + ca) ).Therefore, applying Cauchy-Schwarz, we get:[frac{a}{b + c} + frac{b}{c + a} + frac{c}{a + b} = sum frac{a^2}{ab + ac} geq frac{(a + b + c)^2}{2(ab + bc + ca)}]Okay, so now I have:[frac{(a + b + c)^2}{2(ab + bc + ca)} geq frac{3}{2}]Wait, is that true? Let me check. If I can show that:[frac{(a + b + c)^2}{2(ab + bc + ca)} geq frac{3}{2}]Then, multiplying both sides by ( 2(ab + bc + ca) ), I get:[(a + b + c)^2 geq 3(ab + bc + ca)]Which simplifies to:[a^2 + b^2 + c^2 + 2(ab + bc + ca) geq 3(ab + bc + ca)]Subtracting ( 3(ab + bc + ca) ) from both sides:[a^2 + b^2 + c^2 - ab - bc - ca geq 0]Which is equivalent to:[frac{1}{2} left( (a - b)^2 + (b - c)^2 + (c - a)^2 right) geq 0]Since squares are always non-negative, this inequality holds true. Therefore, the previous step is valid, and thus:[frac{(a + b + c)^2}{2(ab + bc + ca)} geq frac{3}{2}]Which implies:[frac{a}{b + c} + frac{b}{c + a} + frac{c}{a + b} geq frac{3}{2}]So, that's the proof using the Cauchy-Schwarz inequality. But wait, I think I might have skipped a step or two. Let me go through it again to make sure I didn't make any mistakes.Starting from the expression:[frac{a}{b + c} + frac{b}{c + a} + frac{c}{a + b}]I rewrote each term as ( frac{a^2}{ab + ac} ), which is correct because ( frac{a}{b + c} = frac{a^2}{a(b + c)} = frac{a^2}{ab + ac} ). Then, applying the Cauchy-Schwarz inequality in the form of the Titu's lemma, which is a specific case of Cauchy-Schwarz, I get:[sum frac{a^2}{ab + ac} geq frac{(a + b + c)^2}{sum (ab + ac)}]Calculating the denominator, ( sum (ab + ac) = 2(ab + bc + ca) ), which is correct. So, the inequality becomes:[frac{(a + b + c)^2}{2(ab + bc + ca)} geq frac{3}{2}]Multiplying both sides by ( 2(ab + bc + ca) ):[(a + b + c)^2 geq 3(ab + bc + ca)]Expanding the left side:[a^2 + b^2 + c^2 + 2(ab + bc + ca) geq 3(ab + bc + ca)]Subtracting ( 3(ab + bc + ca) ) from both sides:[a^2 + b^2 + c^2 - ab - bc - ca geq 0]Which is indeed true because it can be written as:[frac{1}{2} left( (a - b)^2 + (b - c)^2 + (c - a)^2 right) geq 0]Since squares are always non-negative, the inequality holds. Therefore, the original inequality is proven.Wait, but I think I should also consider if there are other methods to prove this, just to be thorough. Maybe using the AM-HM inequality or some other approach.Let me try using the AM-HM inequality. The AM-HM inequality states that for positive real numbers, the arithmetic mean is greater than or equal to the harmonic mean. But I'm not sure how to apply that directly here. Maybe if I consider the reciprocals or something.Alternatively, I could try the method of substitution. Let me set ( x = b + c ), ( y = c + a ), and ( z = a + b ). Then, the expression becomes:[frac{a}{x} + frac{b}{y} + frac{c}{z}]But I'm not sure if this substitution helps. Maybe I need to express ( a, b, c ) in terms of ( x, y, z ). Let's see:From the substitution:[x = b + c y = c + a z = a + b]Adding all three equations:[x + y + z = 2(a + b + c)]So, ( a + b + c = frac{x + y + z}{2} ). Then, solving for ( a, b, c ):[a = frac{y + z - x}{2} b = frac{x + z - y}{2} c = frac{x + y - z}{2}]Substituting back into the expression:[frac{a}{x} + frac{b}{y} + frac{c}{z} = frac{frac{y + z - x}{2}}{x} + frac{frac{x + z - y}{2}}{y} + frac{frac{x + y - z}{2}}{z}]Simplifying each term:[= frac{y + z - x}{2x} + frac{x + z - y}{2y} + frac{x + y - z}{2z}][= frac{1}{2} left( frac{y}{x} + frac{z}{x} - 1 + frac{x}{y} + frac{z}{y} - 1 + frac{x}{z} + frac{y}{z} - 1 right)][= frac{1}{2} left( left( frac{x}{y} + frac{y}{x} right) + left( frac{x}{z} + frac{z}{x} right) + left( frac{y}{z} + frac{z}{y} right) - 3 right)]Now, I know that for any positive real numbers, ( frac{x}{y} + frac{y}{x} geq 2 ) by AM-GM. Similarly for the other pairs. So, each pair ( frac{x}{y} + frac{y}{x} ), ( frac{x}{z} + frac{z}{x} ), and ( frac{y}{z} + frac{z}{y} ) is at least 2. Therefore, the sum of these three pairs is at least 6.Substituting back:[frac{1}{2} left( text{something} geq 6 - 3 right) = frac{1}{2} times 3 = frac{3}{2}]Wait, that seems a bit hand-wavy. Let me write it more precisely.Since ( frac{x}{y} + frac{y}{x} geq 2 ), ( frac{x}{z} + frac{z}{x} geq 2 ), and ( frac{y}{z} + frac{z}{y} geq 2 ), adding these inequalities gives:[left( frac{x}{y} + frac{y}{x} right) + left( frac{x}{z} + frac{z}{x} right) + left( frac{y}{z} + frac{z}{y} right) geq 6]Therefore, substituting back into the expression:[frac{1}{2} left( text{something} geq 6 - 3 right) = frac{1}{2} times 3 = frac{3}{2}]Wait, no, that's not quite accurate. Let me correct that.The expression inside the brackets is:[left( frac{x}{y} + frac{y}{x} right) + left( frac{x}{z} + frac{z}{x} right) + left( frac{y}{z} + frac{z}{y} right) - 3]We know that the sum of the pairs is at least 6, so:[6 - 3 = 3]Therefore, the entire expression is at least ( frac{1}{2} times 3 = frac{3}{2} ). So, this method also leads us to the conclusion that:[frac{a}{b + c} + frac{b}{c + a} + frac{c}{a + b} geq frac{3}{2}]Okay, so that's another way to prove it using substitution and the AM-GM inequality. It seems like both methods work, but the Cauchy-Schwarz approach is more straightforward.Let me also consider if there's a way to use 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 one is ordered in the opposite way, it's minimized. But I'm not sure how to apply that here directly.Alternatively, maybe I can use the method of Lagrange multipliers to find the minimum of the expression under the constraint ( a + b + c = text{constant} ). But that might be overcomplicating things for this problem.Wait, another thought: perhaps I can use the substitution ( x = b + c ), ( y = c + a ), ( z = a + b ) as before, and then express the inequality in terms of ( x, y, z ). But I think I already tried that and it led to the same conclusion.Let me also check if the inequality holds for specific values. For example, if ( a = b = c ), then each term becomes ( frac{a}{2a} = frac{1}{2} ), so the sum is ( frac{3}{2} ), which matches the lower bound. So, the inequality holds as equality when ( a = b = c ).What if one variable is much larger than the others? Let's say ( a ) is very large compared to ( b ) and ( c ). Then, ( frac{a}{b + c} ) would be very large, while the other terms ( frac{b}{c + a} ) and ( frac{c}{a + b} ) would be small. But since ( a ) is positive, the sum would still be greater than ( frac{3}{2} ). Wait, actually, if ( a ) is very large, ( frac{a}{b + c} ) would dominate, making the sum much larger than ( frac{3}{2} ). So, the inequality still holds.What if one variable approaches zero? Let's say ( c ) approaches zero. Then, the expression becomes ( frac{a}{b} + frac{b}{a} + 0 ). The sum ( frac{a}{b} + frac{b}{a} ) is always at least 2 by AM-GM, which is greater than ( frac{3}{2} ). So, the inequality holds in this case as well.Therefore, the inequality seems to hold in various cases, which gives me more confidence in the proof.In summary, the most straightforward method is to apply the Cauchy-Schwarz inequality (or Titu's lemma) to the given expression, which leads directly to the desired result. The substitution method also works but involves more steps and relies on recognizing the structure of the expression.I think I've covered the main approaches and verified the inequality through different methods and specific cases. So, I'm confident that the proof is correct.