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:
$xyz\leq 1$.
 A: These inequalities are algebraic and thus can be proved purely algorithmically.
Mathematica takes a minute or two for this proof of your first inequality:


Here is a "more human" proof:
Substituting $z=\frac1{xy}$, rewrite your first inequality as
\begin{equation}
    f(x,y)\mathrel{:=}\left(\frac{1}{x^{10} y^{10}}+x^{10}+y^{10}\right)^2-3 \left(\frac{1}{x^{13} y^{13}}+x^{13}+y^{13}\right)\ge0
\end{equation}
and then as
\begin{align}
g(x,y)&\mathrel{:=}f(x,y)x^{20} y^{20} \\ 
&=x^{40} y^{20}-3 x^{33} y^{20}+2 x^{30} y^{30}+x^{20} y^{40}-3 x^{20} y^{33} \\ 
&+2 x^{20} y^{10}+2 x^{10}
   y^{20}-3 x^7 y^7+1\ge0,
\end{align}
for $x,y>0$.
Further,
\begin{align}
g_1(x,y)\mathrel{:=}{}\frac{g'_x(x,y)}{x^6 y^7}&=40 x^{33} y^{13}-99 x^{26} y^{13}+60 x^{23} y^{23}+20 x^{13} y^{33} \\ 
&-60 x^{13} y^{26}+40 x^{13} y^3+20 x^3
   y^{13}-21, \\
g_2(x,y)\mathrel{:=}\frac{g'_y(x,y)}{x^7 y^6}&=20 x^{33} y^{13}-60 x^{26} y^{13}+60 x^{23} y^{23}+40 x^{13} y^{33} \\ 
&-99 x^{13} y^{26}+20 x^{13} y^3+40 x^3
   y^{13}-21,
\end{align}
and the only positive roots of the resultants of $g_1(x,y)$ and $g_2(x,y)$ with respect to $x$ and $y$ are $y=1$ and $x=1$, respectively. So, $(1,1)$ is the only critical point of $g$.
Next, all the coefficients of the polynomial $g(1+u,1+v)$ in $u,v$ are nonnegative. Therefore and because of the symmetry $x\leftrightarrow y$, it remains to consider the cases (i) $0\le x\le1$ and $y>0$ is large enough and (ii) $x=0$.
For case (i), we have $g(x,y)\ge1 - 3 x^7 y^7 + 2 x^{10} y^{20}>0$. For case (ii), we have $g(0,y)=1>0$.
So, your first inequality is proved, again.

It took Mathematica about 1.8 hours to prove your second inequality (click on the image to enlarge it):

The latter proof would probably take many thousands pages.
A: Here is a little less computer-assisted approach to both inequalities than the one suggested by Iosif Pinelis. Namely, I use the properties of rational one-variable functions which are seen from their graphs without thinking about how to prove them rigorously.

*

*Fix $xyz=1$ and $x^{10}+y^{10}+z^{10}:=S$. Look for a maximum of $x^{13}+y^{13}+z^{13}$. It is achieved (the set of admissible triples is compact), and at the maximum points the gradients of $xyz,x^{10}+y^{10}+z^{10},x^{13}+y^{13}+z^{13}$ are linearly dependent, that is, we should have $$\alpha\cdot(yz,xz,xy)+\beta\cdot 10(x^9,y^9,z^9)+\gamma\cdot 13(x^{12},y^{12},z^{12})=0$$
where the real coefficients $(\alpha,\beta,\gamma)$ are not simultaneously zero. This means that all numbers $x,y,z$ solve the same equation $f(t):=a+bt^{10}+ct^{13}=0$, where $a=\alpha xyz$, $b=10\beta$, $c=13\gamma$. Such an equation may have at most two different positive solutions by Descartes' rule of signs. That is, two of $x,y,z$ must be equal. Without loss of generality we may assume that $y=x$, then $z=1/x^2$ and we should prove a 1-variable inequality $$(2x^{10}+x^{-20})^2\geqslant 3(2x^{13}+x^{-26}).$$
I do not see any nice explanation why this is true, but looking at the graphs we see that the ratio of exponents 13:10 may be increased to approximately 1.4047 (you may see that this graph is above the x-axis, but for the value of parameter 1.4048 it crosses it already.


*Denote $a=y/x,b=z/y,c=x/z$. Then $abc=1$ and we are given $(xyz)^4(2+3a^4)(2+3b^4)(2+3c^4)\leqslant xyz(3+2a)(3+2b)(3+2c)$. Thus for establishing $xyz\leqslant 1$ it suffices (and is actually necessary) to check that $$(3+2a)(3+2b)(3+2c)\leqslant (2+3a^4)(2+3b^4)(2+3c^4) \quad (1)$$
whenever $a,b,c$ are positive numbers with $abc=1$. (1) is equivalent to
$$
F(a,b,c):=h(a)+h(b)+h(c)\geqslant 0, \,\, \text{where}\,\, h(x):=\log(2+3x^4)-\log(3+2x).
$$
First of all, I claim that $F$ attains its minimal value on the set $\Omega:=\{(a,b,c):abc=1, a,b,c>0\}$. Indeed, consider a sequence $(a_n,b_n,c_n)$ for which $F(a_n,b_n,c_n)$ approaches the infimum of $F$ on $\Omega$. Since $h(x)$ is bounded from below on $(0,\infty)$ and tends to $+\infty$ for large $x$, we conclude that $a_n,b_n,c_n$ must be bounded, then we may choose a convergent subsequence and get a minimizer. So, denote the minimizer by $(a_0,b_0,c_0)$. the gradients of $F(a,b,c)$ and $abc$ at the point $(a_0,b_0,c_0)$ must be linearly dependent, thus we may write $h'(a_0)=\lambda b_0c_0$, $h'(b_0)=\lambda a_0c_0$, $h'(c_0)=\lambda b_0a_0$ for certain real $\lambda$. In other words, the function $g(x):=xh'(x)$ takes the same value at points $a_0,b_0,c_0$. Looking at the plot of $g(x)$ for positive $x$ we see that it takes each value at most twice. Thus two of three variables $a_0,b_0,c_0$ must be equal. Without loss of generality $a_0=b_0=:x$, $c_0=1/x^2$, and we should prove a 1-variable polynomial inequality $$(2+3x^4)^2(2+3/x^8)\geqslant (3+2x)^2(3+2/x^2)\quad \text{for}\quad x>0.$$
Well, it follows from factorization.
A: Another way. By my previous post it's enough to prove that:
$$(x^5+y^5+z^5)^2\geq3xyz(x^7+y^7+z^7)$$ for positive $x$, $y$ and $z$.
Indeed, let $x^5+y^5+z^5=\text{constant}$ and $x^7+y^7+z^7=\text{constant}$.
Thus, by the Vasc's EV Method (see here: Cîrtoaje - The equal variable method Corollary 1.8(b)) it's enough to prove the last inequality (because it's homogeneous) for $y=z=1$, which gives
$$(x^5+2)^2\geq3x(x^7+2)$$ or
$$(x-1)^2(x^8+2x^7-2x^5-4x^4-2x^3+2x+4)\geq0$$ and the rest is smooth.
A: This is to complement the nice answer by Fedor Petrov by a calculus proof of 
the inequality
$$ (2 x^{10} + x^{-20})^2\ge3 (2 x^{13} + x^{-26}),$$
for real $x>0$.
Rewrite this inequality as
$$f(x):=4 x^{60}-6 x^{53}+4 x^{30}-3 x^{14}+1\ge0.$$
Let
$$f_1(x):=\frac{f'(x)}{6x^{13}}=40 x^{46}-53 x^{39}+20 x^{16}-7,\\f_2(x):=\frac{f_1'(x)}{x^{15}}=1840 x^{30}-2067 x^{23}+320,\\
f_3(x):=\frac{f_2'(x)}{69x^{22}}=800 x^7-689.$$
Then, clearly, $f_2$ attains its minimum (on $[0,\infty)$) at $(689/800)^{1/7}$, and this minimum is $24.7\ldots>0$. So, $f_2>0$ and hence $f_1$ is increasing, from $f_1(0)=-7<0$ to $f_1(\infty-)=\infty>0$. Also, $f_1(1)=0$. So, $f_1\le0$ on $[0,1]$ and $f_1\ge0$ on $[1,\infty)$. So, $f$ attains its minimum (on $[0,\infty)$) at $1$, and this minimum is $0$.
A: 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.
