This is Problem 12024 of AMM.
It asks to show that if $x,y,z$ are positive reals, and $xyz=1$, then $(x^{10}+y^{10}+z^{10})^{2}\geq 3(x^{13}+y^{13}+z^{13})$.
I could show it for the particular case of $z=1$, when it becomes a one variable polynomial inequality.
For the general problem, I thought about homogenizing, by multiplying the right side by $(xyz)^{\frac{7}{3}}$, but couldn't make progress.
Best Answer
We'll prove that $$\left(a^{10}+b^{10}+c^{10}\right)^2\geq3(a^{14}+b^{14}+c^{14}),$$ where $abc=1,$ for which it's enough to prove that $$(x^5+y^5+z^5)^2\geq3xyz(x^7+y^7+z^7)$$ for positives $x$, $y$ and $z$.
Indeed, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Hence, $$\left(x^5 + y^5 + z^5\right)^2-3xyz\left(x^7 +y^7 + z^7\right)=$$ $$=(u^2-uv+v^2)x^8+8(-u^3+2u^2v+2uv^2-v^3)x^7+$$ $$+4(5u^4-17u^3v+50u^2v^2-17uv^3+5v^4)x^6+$$ $$+8(11u^5-20u^4v+25u^2v^2+25u^2v^3-20uv^4+11v^5)x^5+$$ $$+2(63u^6-79u^5v+50u^4v^2+100u^3v^3+50u^2v^4-79uv^5+63u^6)x^4+$$ $$+4(24u^7-21u^6v+5u^5v^2+25u^4v^3+25u^3v^4+5u^2v^5-21uv^6+24v^7)x^3+$$ $$+2(21u^8-12u^7v+10u^5v^3+25u^4v^4+10u^3v^5-12uv^7+21v^8)x^2+$$ $$+(10u^9-3u^8v+10u^5v^4+10u^4v^5-3uv^8+10v^9)x+(u^5+v^5)^2\geq$$ $$\geq\left((u^2-uv+v^2)x^2-8(u+v)(u^2-3uv+v^2)x+4(5u^4-17u^3v+50u^2v^2-17uv^3+5v^4)\right)x^6.$$
Let $u^2+v^2=tuv$.
Hence, $t\geq2$ and it remains to prove that $$(t-1)(5t^2-17t+40)-4(t+2)(t-3)^2\geq0,$$ which is true because $$(t-1)(5t^2-17t+40)-4(t+2)(t-3)^2=$$ $$=t^3-6t^2+69t-112=t(t-3)^2+60t-112\geq0.$$ Id est, it's enough to prove that $$x^{14}+y^{14}+z^{14}\geq x^{13}+y^{13}+z^{13},$$ where $x$, $y$ and $z$ are positives such that $xyz=1$, or $$x^{42}+y^{42}+z^{42}\geq (x^{39}+y^{39}+z^{39})xyz,$$ which is true by Muirhed because $$(42,0,0)\succ(40,1,1).$$