# Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{a^3+2b^3}+\frac{b^3}{b^3+2c^3}+\frac{c^3}{c^3+2a^3} \geq 1.$

Prove that if $a,b,$ and $c$ are positive real numbers, then $$\dfrac{a^3}{a^3+2b^3}+\dfrac{b^3}{b^3+2c^3}+\dfrac{c^3}{c^3+2a^3} \geq 1.$$

This question seems hard since we aren't given any other information about $a,b,c$. I think expanding it might got somewhere, but I don't see Cauchy-Schwarz or AM-GM leading anywhere.

Let $x=a^3$, $y=b^3$, $z=c^3$. By Cauchy-Schwarz inequality:

$$\left(\sum_{\text{cyc}}\frac{x}{x+2y}\right)\ge \frac{(x+y+z)^2}{\left(\sum_{\text{cyc}} x(x+2y)\right)}=1$$

Equality holds iff $\frac{\frac{x}{x+2y}}{x(x+2y)}=\frac{\frac{y}{y+2z}}{y(y+2z)}=\frac{\frac{z}{z+2x}}{z(z+2x)}$, i.e. iff $x+2y=y+2z=z+2x$,

i.e. iff $y=2z-x=2x-z$, i.e. iff $x=y=z$, i.e. iff $a=b=c$.

It's quite a standard trick for similar inequalities.

First, notice that $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$, because this is equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true (or by the Rearrangement Inequality).

E.g., if $a,b,c>0$ and $n\in\mathbb Z^+$, then:

$$\sum_{\text{cyc}}\frac{a}{b+nc}\ge \frac{(a+b+c)^2}{\sum_{\text{cyc}}a(b+nc)}=\frac{\sum_{\text{cyc}}a^2+2\sum_{\text{cyc}}ab}{(n+1)\sum_{\text{cyc}}ab}$$

$$\ge \frac{\sum_{\text{cyc}}ab+2\sum_{\text{cyc}}ab}{(n+1)\sum_{\text{cyc}}ab}=\frac{3}{n+1}$$

$n=1$ gives Nesbitt's Inequality.

• All to easy. +1 Well done. – Mark Viola Jan 10 '16 at 21:56
• @Dr.MV I've added an application of this trick to another inequality. – user236182 Jan 10 '16 at 22:19
• That's a nice inequality. – Mark Viola Jan 10 '16 at 22:48