<s>[INST] How to prove this high-degree inequality Let $x$,$y$,$z$ be positive real numbers which satisfy $xyz=1$. Prove that:
$(x^{10}+y^{10}+z^{10})^2 \geq 3(x^{13}+y^{13}+z^{13})$.
And there is a similar question: Let $x$,$y$,$z$ be positive real numbers which satisfy the inequality
$(2x^4+3y^4)(2y^4+3z^4)(2z^4+3x^4) \leq(3x+2y)(3y+2z)(3z+2x)$. Prove this inequality:
$xyz\leq 1$.
 [/INST] This is to give an alternative proof of the inequality in Fedor Petrov's nice answer
$$(2 x^{10} + x^{-20})^2\ge3 (2 x^{13} + x^{-26})$$
for all $x > 0$.
We have
\begin{align*}
    &2x^{13+1/3} + x^{-26-2/3} - (2x^{13} + x^{-26})\\
    ={}& 2(x^{13+1/3} - x^{13}) + (x^{-13-1/3} - x^{-13})
    (x^{-13-1/3} + x^{-13})\\
    ={}& x^{13}(x^{1/3}-1)(2 - x^{-119/3} - x^{-118/3})\\
    \ge{}& 0.
\end{align*}
Thus, it suffices to prove that
$$(2 x^{10} + x^{-20})^2\ge 3(2x^{13+1/3} + x^{-26-2/3}).$$
Letting $x = a^{3/10}$, it suffices to prove that, for all $a > 0$,
$$(2a^3 + a^{-6})^2 \ge 3(2a^4 + a^{-8}).$$
We have
\begin{align*}
    &(2a^3 + a^{-6})^2 - 3(2a^4 + a^{-8})\\
    ={}& 4a^6 + 4a^{-3}+ a^{-12} - 6a^4 - 3a^{-8}\\
    \ge\,& 4a^6 + 2(3a^{-2} - 1) + a^{-12}  - 6a^4 - 3a^{-8}\\
    ={}& \frac{(2a^2+1)(2a^{12} - 2a^6 + 1)(a^2-1)^2}{a^{12}}\\
    \ge{}& 0
\end{align*}
where we have used
$2a^{-3} - (3a^{-2} - 1) = \frac{(a+2)(a-1)^2}{a^3} \ge 0$.
We are done.
</s>