<s>[INST] $x+y+z=3$, prove the inequality For $x,y,z>0$ and $x+y+z=3$ prove that $\frac{x^3}{(y+2z)^2}+\frac{y^3}{(z+2x)^2}+\frac{z^3}{(x+2y)^2}\ge \frac{1}{3}$.
QM, AM, GM, HM suggested ;)
 [/INST] Use the inequality $\dfrac{a^3}{b^2}\geq \dfrac{2a^2}{3b} - \dfrac{a}{9}$ and then use $\dfrac{a^2}{b}\geq \dfrac{2a}{3}-\dfrac{b}{9}$, so $$\sum_{x,y,z}\dfrac{x^3}{(y+2z)^2}\geq\sum_{x,y,z}(\dfrac{x}{3} - \dfrac{2(y+2z)}{27})=\dfrac{3(x+y+z)}{27}=\dfrac{1}{3}.$$
The inequalities used are trivial AM-GM.  
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