Prove the Inequality: $\sum\frac{x^3}{2x^2+y^2}\ge\frac{x+y+z}{3}$ Let $x, y, z&gt;0$. Prove that:
$$\frac{x^3}{2x^2+y^2}+\frac{y^3}{2y^2+z^2}+\frac{z^3}{2z^2+x^2}\ge\frac{x+y+z}{3}$$
See my solution from 2006  here:

https://artofproblemsolving.com/community/c6h22937p427220
I have a direction which may be fruitful but didn't work out all cases. Rewrite the inequality as:
$$
\frac{x^3}{2x^2+y^2}-\frac{x}{3}+\frac{y^3}{2y^2+z^2}-\frac{y}{3}+\frac{z^3}{2z^2+x^2}-\frac{z}{3}\ge0
$$

Let $f(t)=\frac{1-t^2}{2+t^2}$, then the inequality is equivalent to: $xf(a)+yf(b)+zf(c)\ge0$, where $a=y/x, b=z/y $ and $c=x/z$.

Case 1. Assume that $a,b,c \in (0,2]$. Note that $f(t)\ge \frac{1-t}{2}$ for each $t \in (0,2]$. Then we have $$xf(a)+yf(b)+zf(c) \ge \frac{x}{2}(1-\frac{y}{x})+\frac{y}{2}(1-\frac{z}{y})+\frac{z}{2}(1-\frac{x}{z})=0$$

Case 1 is settled.

Case 2 $x\ge y\ge z$ and $x\ge 2z$ remains to be settled.

Case 3 $z\ge y\ge x$ and $z\ge 2y$ OR $y\ge 2x$ remains to be settled.