Multiple equality conditions

Prove that if $a, b, c$ are non-negative real numbers such that $a + b + c = 3$ , then we have

$ab^2 + bc^2 + ca^2 + abc \leq 4 .$

#Algebra

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Comments

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 $f(a, b, 0) \leq 4 = f ( 1, 2, 0)$ .

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?

Calvin Lin Staff - 6 years, 1 month ago

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

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

Prasun Biswas - 6 years, 1 month ago

Thanks fixed.

Calvin Lin Staff - 6 years, 1 month ago

Sir, What is meant by IE ?

Priyanshu Mishra - 5 years, 7 months ago

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

Prasun Biswas - 5 years, 7 months ago

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:

If $a,b,c$ are non-negative reals, then the following inequality holds:

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

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.

Prasun Biswas - 6 years, 1 month ago

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

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

Calvin Lin Staff - 6 years, 1 month ago

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

Krishna Sharma - 6 years, 1 month ago

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

Calvin Lin Staff - 6 years, 1 month ago

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