# Prove the follwing inequality : (x^2y^2+z^4)3<=(x^3+y^3+z^3)^4

Let $x,y,z$ be three positive numbers. Can you prove the follwing inequality : $\left({x}^{2}{y}^{2}+{z}^{4}{\right)}^{3}\le \left({x}^{3}+{y}^{3}+{z}^{3}{\right)}^{4}$
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eri1ti0m
We need to show $\left({x}^{2}{y}^{2}+{z}^{4}{\right)}^{3}\le \left({x}^{3}+{y}^{3}+{z}^{3}{\right)}^{4}.$
By AM-GM, we have $\left({x}^{3}+{y}^{3}+{z}^{3}{\right)}^{4}\ge {\left(2\left(xy{\right)}^{\frac{3}{2}}+{z}^{3}\right)}^{4}$
Let $a=\sqrt{xy}>0$. Then it is sufficient to show that$\left(2{a}^{3}+{z}^{3}{\right)}^{4}\ge \left({a}^{4}+{z}^{4}{\right)}^{3}$
Let $t=\frac{a}{z}>0$, then we need to show
$f\left(t\right)=\left(2{t}^{3}+1{\right)}^{4}-\left({t}^{4}+1{\right)}^{3}\ge 0$ for $t>0$
or $f\left(t\right)={t}^{3}\left(15{t}^{9}+32{t}^{6}-3{t}^{5}+24{t}^{3}-3t+8\right)>0$
Now, note that for $t\ge 1$
$3{t}^{6}-3{t}^{5}=3{t}^{5}\left(t-1\right)$ and $3{t}^{3}-3t=3t\left({t}^{2}-1\right)$
so $f\left(t\right)={t}^{3}\left[15{t}^{9}+32{t}^{6}+3{t}^{3}\left(1-{t}^{2}\right)+21{t}^{3}+3\left(1-t\right)+5\right]>0.$
Thus $f\left(t\right)>0$ when t>0, and hence the inequality holds.