Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$

Question

Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.

My (strange) proof:

$$ \begin{align*} a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\ \sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\ a^2+b^2+c^2 &\geq ab+bc+ca\\ 2a^2+2b^2+2c^2-2ab-2bc-2ca &\geq 0\\ \left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &\geq 0 \end{align*} $$

Which is obviously true.

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However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:

$$ \begin{align*} \sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\ a+b+c &\geq a+b+c \end{align*} $$

Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?

Edit: Please don't help me with the original inequality, I'll figure it out.

Answer 1

Just assume, wlog $a\leq b\leq c$. Then this equation is all you need: $$a^3+b^3+c^3=a^2b+b^2c+c^2a+\underset{\geq 0}{\underbrace{(c^2-a^2)(b-a)}}+\underset{\geq 0}{\underbrace{(c^2-b^2)(c-b)}}\geq a^2b+b^2c+c^2a$$

Answer 2

WOLG, Let $c$=Max{$a,b,c$}, then there is 2 cases:

case I: $0<a \le b \le c$, we want to prove $c^2(c-a) \ge a^2(b-a)+b^2(c-b)$

we have $c^2\ge b^2, c^2\ge a^2 \to $,RHS $\le c^2(b-a)+c^2(c-b)=c^2(c-a)$

case II: $0<b \le a \le c$, we want to prove $a^2(a-b)+c^2(c-a) \ge b^2(c-b)$

we have $a^2\ge b^2,c^2 \ge b^2, \to$LHS $ \ge b^2(a-b)+b^2(c-a)=b^2(c-b)$

to summary 2 cases, we have $a^2(a-b)+b^2(b-c)+c^2(c-a) \ge 0$

QED

Answer 3

Without making any assumption, just simple $AM\ge GM$ $$a^3+a^3+b^3\ge3a^2b$$ $$b^3+b^3+c^3\ge3b^2c$$ $$c^3+c^3+a^3\ge3c^2a$$ $$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$$