Half-symmetric, homogeneous inequality Let $x,y,z$ be three positive numbers.  Can anybode prove the follwing inequality :
$(x^2y^2+z^4)^3 \leq (x^3+y^3+z^3)^4$ (or find a counterexample, or find a reference ...)
 A: We need to show $(x^2y^2 + z^4)^3 \le (x^3+y^3+z^3)^4.$  
By AM-GM, we have $(x^3+y^3+z^3)^4 \ge \left(2(xy)^{\frac{3}{2}} + z^3\right)^4$
Let $a = \sqrt{xy} > 0$.  Then it is sufficient to show that
$(2a^3 + z^3)^4 \ge (a^4 + z^4)^3$
Let $t = \frac{a}{z} > 0$, then we need to show
$f(t) = (2t^3 + 1)^4 - (t^4 + 1)^3 \ge 0$ for $t > 0$.
or $f(t) = t^3 (15 t^9+32 t^6-3 t^5+24 t^3-3t+8) > 0$
Now, note that for $t \ge 1$,
$3t^6 - 3t^5 = 3t^5(t-1)$  and $3t^3 - 3t = 3t(t^2 - 1)$
so $f(t) = t^3 [15t^9 + 29t^6 + 3t^5(t-1) + 21t^3 + 3t(t^2-1) + 8] > 0$. 
Similarly, when $0 < t < 1$,
$- 3t^5 + 3t^3 = 3t^3(1 - t^2)$  and $-3t + 3 = 3(1-t)$
so $f(t) = t^3 [15t^9 + 32t^6 + 3t^3(1-t^2) + 21t^3 + 3(1-t) + 5] > 0.$  
Thus $f(t) > 0$ when $t > 0$, and hence the inequality holds.
A: Note that we have $(x^3+y^3+z^3)^2\geq (x^3+y^3)^2+z^6\geq 4x^3y^3+z^6,$ and also $x^3+y^3\geq 2\sqrt{x^3y^3},$ so that it suffices to check that
$$\left(2\sqrt{x^3y^3}+z^3\right)^2(4x^3y^3+z^6)\geq (x^2y^2+z^4)^3.$$
Using the Holder's inequality, we can get a sharper bound:
$$\begin{aligned}\left(2\sqrt{x^3y^3}+z^3\right)^2(4x^3y^3+z^6)&\geq \left(\sqrt[3]{16x^6y^6}+z^4\right)^3\\&=\left(2\sqrt[3]2x^2y^2+z^4\right)^3.\end{aligned}$$
In this stronger version, equality holds for $(x,y,z)=(t,t,\sqrt[3]{2t}).$
For the original inequality, if nonnegative reals are allowed, then equality holds if and only if $x=y=0.$ Otherwise there is no equality.
$\Box$
