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This is quite a famous inequality, and there are several approaches that could be taken. Here is the simplest proof that I know of.

Define $f( a, b, c) = ab^2 + bc^2 + ca^2 + abc$.
WLOG, we may assume that $a$ is the median of the 3, IE we either have $b \leq a \leq c$ or $b \geq a \geq c$.

Step 1: We will show that $f(a,b,c) \leq f ( a, b+c, 0 )$. This approach is known as Smoothing.

This follows by expanding both sides, and we want to compare $ab^2 + bc^2 + ca^2 + abc \leq a(b+c)^2$, which simplifies to

$bc^2 + ca^2 \leq abc + ac^2$

This is equivalent to

$c ( a - b ) ( a - c) \leq 0$

From the assumption that $a$ is the middle number, we know that $a-b, a-c$ will have different signs (or be 0). Since $c$ is non-negative, hence the entire product will be $\leq 0$.

This is essentially how we approach @Krishna Sharma 's Case 1.

Step 2: We show that subject to $a + b = 3$, we have $f(a, b, 0) \leq 4 = f ( 1, 2, 0)$.

Our required inequality trivially follows from the stated inequality. However, proving the said stronger inequality seems to be harder than I expected. Let me see if I can think of a proof to that. For the time being, others are welcome to post their proof (if any) for the stated stronger inequality.

Note that the inequalities are equivalent. What happened was that we normalized the inequality, meaning that we made all of the terms have the same polynomial degree. To do so, we multiplied, where necessary, by $a + b + c = 3$.

This is a standard approach, and one that I would often (though not always) recommend to use if the terms have different degrees.

The weaker version that you saw, was most likely

$ab^2 + bc^2 + ca^2 \leq 4 \Leftrightarrow ab^2 + bc^2 + ca^2 \leq \frac{4}{27} ( a + b + c ) ^ 3 .$

You did not perform the Lagrangian properly. At the IMO, this solution will be scored 0/7.
1. You did not state the equations.
2. You did not state how to solve the equations.
3. You missed out the equality case of $(1, 2, 0)$.

@Aman Rajput
–
You missed it because you didn't take care of the boundary condition. IE You didn't perform the Lagrangian properly.

E.g. What is the maximum of $f(x) = x^2$ on the interval $[-2, 2]$? Do you say that " $f' = 0 \Rightarrow x = 0$ hence the maximum is $f(0) = 0$? No, we still have to check the boundary points, where they need not satisfy $f' = 0$ in order to be a maximum on the restricted domain.

@Calvin Lin
–
I know what you are trying to say . But what i know is that the min / max will be obtained using lagrangian even if we missed out boundary condition or other ordered pairs of equality

For example, if you ignore the boundary condition when calculating the max of $f( x) = x^2$ on the closed interval $[1, -1]$, then you would not get any answer.

This problem becomes quite a standard exercise when you use Lagrange multipliers. I think that the challenge is to prove the given inequality without using that which would be the reason for the problem being tagged under Algebra.

@Aman Rajput
–
What Prasun said is actually true. Lagrange Multiplier makes this problem too easy. It's like you're using a chainsaw to cut a carrot.

On the other hand, you need to show that the extremal point you've found is a global maximum point as opposed to an inflection point or a global minimum point.

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## Comments

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TopNewestThis is quite a famous inequality, and there are several approaches that could be taken. Here is the simplest proof that I know of.

Define $f( a, b, c) = ab^2 + bc^2 + ca^2 + abc$. WLOG, we may assume that $a$ is the median of the 3, IE we either have $b \leq a \leq c$ or $b \geq a \geq c$.

Step 1:We will show that $f(a,b,c) \leq f ( a, b+c, 0 )$. This approach is known as Smoothing.This follows by expanding both sides, and we want to compare $ab^2 + bc^2 + ca^2 + abc \leq a(b+c)^2$, which simplifies to

$bc^2 + ca^2 \leq abc + ac^2$

This is equivalent to

$c ( a - b ) ( a - c) \leq 0$

From the assumption that $a$ is the middle number, we know that $a-b, a-c$ will have different signs (or be 0). Since $c$ is non-negative, hence the entire product will be $\leq 0$.

This is essentially how we approach @Krishna Sharma 's Case 1.

Step 2:We show that subject to $a + b = 3$, we have $f(a, b, 0) \leq 4 = f ( 1, 2, 0)$.By AM-GM, we get

$f(a, b, 0) = ab^2 = 4 \times a \times \frac{b}{2} \times \frac{b}{2} \leq 4 \left( \frac{a + \frac{b}{2} + \frac{b}{2} } { 3} \right) ^3 = 4 .$

Hence, the result follows.

However, I do not know of an easy way to motivate the approach, and in particular step 1. Any thoughts or comments?

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Minor typo in the last step where you use AM-GM.

$4\times a\times \left(\frac{b}{2}\right)^2\leq 4\left(\frac{a+\frac{b}{2}+\frac{b}{2}}{\color{#D61F06}{3}}\right)^3=4$

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Thanks fixed.

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Sir, What is meant by IE ?

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It's just another way to write i.e. which means "that is".

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This is just a comment, not a solution.I remember seeing a stronger version of this inequality somewhere recently (probably on Math SE) which stated the following:

Our required inequality trivially follows from the stated inequality. However, proving the said

strongerinequality seems to be harder than I expected. Let me see if I can think of a proof to that. For the time being, others are welcome to post their proof (if any) for the statedstrongerinequality.Log in to reply

Note that the inequalities are equivalent. What happened was that we

normalizedthe inequality, meaning that we made all of the terms have the same polynomial degree. To do so, we multiplied, where necessary, by $a + b + c = 3$.This is a standard approach, and one that I would often (though not always) recommend to use if the terms have different degrees.

The weaker version that you saw, was most likely

$ab^2 + bc^2 + ca^2 \leq 4 \Leftrightarrow ab^2 + bc^2 + ca^2 \leq \frac{4}{27} ( a + b + c ) ^ 3 .$

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Is some info missing? Because substituting $a=b=c = \frac{4}{3}$ yields a value around 9.5

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Ooops, the condition should have been $a+b+ c = 3$.

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Let $f(a,b,c)=ab^2+bc^2+ca^2+abc + k(a+b+c-3)$ (say this as equation $(1)$

Solving these four equations $\frac{\partial f}{\partial a} = 0$ $\frac{\partial f}{\partial b} = 0$ $\frac{\partial f}{\partial c} = 0$ $a+b+c=3$

we get $a=b=c=1 , k=-4$

substituting this back in equation $1$ , we get $f_{max}(a,b,c)=1+1+1+1-4(0) = 4$

$\textbf{Q.E.D}$

@Calvin Lin sir

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This demonstrates the point of this problem.

You did not perform the Lagrangian properly. At the IMO, this solution will be scored 0/7.

1. You did not state the equations.

2. You did not state how to solve the equations.

3. You missed out the equality case of $(1, 2, 0)$.

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I didn't miss the case 1,2,0 doesnt satisfy the third equation ,i.e. , df/dc = 0

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boundary condition. IE You didn't perform the Lagrangian properly.E.g. What is the maximum of $f(x) = x^2$ on the interval $[-2, 2]$? Do you say that " $f' = 0 \Rightarrow x = 0$ hence the maximum is $f(0) = 0$? No, we still have to check the boundary points, where they need not satisfy $f' = 0$ in order to be a maximum on the restricted domain.

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For example, if you ignore the boundary condition when calculating the max of $f( x) = x^2$ on the closed interval $[1, -1]$, then you would not get any answer.

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ok i will keep in mind always to check at boundary

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This problem becomes quite a standard exercise when you use Lagrange multipliers. I think that the challenge is to prove the given inequality without using that which would be the reason for the problem being tagged under Algebra.

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i dont think so ... :)

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On the other hand, you need to show that the extremal point you've found is a global maximum point as opposed to an inflection point or a global minimum point.

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Why not assume a=b= c=1 and do it

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Myth: An expression attains it's maximum or minimum when all (some) of the variables are equal.

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