Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.

Solution 1:

First I'll prove a Lemma: $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$.

Proof: it follows from the Rearrangement Inequality, because $(a,b,c)$ and $(a,b,c)$ are similarly sorted.

Or notice that it's equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true; or add the following inequalities: $a^2+b^2\ge 2ab$, $b^2+c^2\ge 2bc$, $c^2+a^2\ge 2ca$. $\ \square$

Your inequality is cyclic. Wlog there are two cases:

• $a\ge b\ge c$. Then $(a,b,c)$ and $(1/a,1/b,1/c)$ are oppositely sorted, so

  $$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{a^3}{a}+\frac{b^3}{b}+\frac{c^3}{c}=a^2+b^2+c^2$$ Now use the Lemma.

• $a\ge c\ge b$. Then $(a,c,b)$ and $(1/a,1/c,1/b)$ are oppositely sorted, so $$\frac{a^3}{b}+\frac{c^3}{a}+\frac{b^3}{c}\ge \frac{a^3}{a}+\frac{c^3}{c}+\frac{b^3}{b}=a^2+c^2+b^2$$ Now use the Lemma.

In your inequality, equality holds if and only if $a=b=c$.

Another proof: By Hölder's inequality:

$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{(a+b+c)^3}{3(a+b+c)}=\frac{(a+b+c)^2}{3}$$

Also $(a+b+c)^2\ge 3(ab+bc+ca)$, because this is equivalent to $a^2+b^2+c^2\ge ab+bc+ca$ (see the Lemma).

Solution 2:

For positive $x$, $y$, $z$, $a$, $b$, and $c$, note that that \begin{align*} (x+y+z)^2 &= \left(\frac{x}{\sqrt{a}}\sqrt{a} + \frac{y}{\sqrt{b}}\sqrt{b} +\frac{z}{\sqrt{c}}\sqrt{c}\right)^2\\ &\le \left(\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c}\right)(a+b+c). \end{align*} That is, \begin{align*} \frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}. \end{align*} Then \begin{align*} \frac{a^3}{b} + \frac{b^3}{c} +\frac{c^3}{a} &=\frac{a^4}{ab} + \frac{b^4}{bc} +\frac{c^4}{ac}\\ &\ge \frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\\ &\ge \frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}\\ &=ab+bc+ac. \end{align*}