A cyclic inequality of degree 10

Question

Suppose that $x,y,z\geq 0$. I would like to prove that $$(x^5+y^5+z^5)^2\geq (x+y+z)(x^3y^6+y^3z^6+z^3x^6).$$ I can prove this inequality using some standard methods. For example, I can let $x=1, y=1+p$, and $z=1+p+q$. Then Mathematica can simplify the difference of the two sides of the inequality to show that all of the coefficients are positive. I don't like this solution, since it relies heavily on a complicated algebraic simplification. Do you have a better proof?

Answer 1

Another way:

After homogenization we need to prove that: $$\sum_{cyc}(x^{10}+2x^5y^5-x^7z^3-x^6z^4-x^6z^3y)\geq0$$ or $$\sum_{cyc}(x^7z^3-x^6z^3y)+\sum_{cyc}(x^{10}-2x^7z^3-x^6z^4+2x^5y^5)\geq0,$$ which is true by AM-GM and SOS: $$\sum_{cyc}(x^7z^3-x^6z^3y)+\sum_{cyc}(x^{10}-2x^7z^3-x^6z^4+2x^5y^5)=$$ $$=\frac{1}{37}\sum_{cyc}(30x^7z^3+4y^7x^3+3z^7y^3-37x^6z^3y)+\sum_{cyc}(x^{10}+2x^5y^5-x^4y^6-2x^3y^7)\geq$$ $$\geq\sum_{cyc}\left(\sqrt[37]{\left(x^7z^3\right)^{30}\left(y^7x^3\right)^4\left(z^7y^3\right)^3}-x^6z^3y\right)+$$ $$=\sum_{cyc}x^3(x-y)(x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+3xy^5+2y^6)=$$ $$=\sum_{cyc}\left(x^3(x-y)(x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+3xy^5+2y^6)-(x^{10}-y^{10})\right)=$$ $$=\sum_{cyc}y^5(x-y)^2(2x^3+3x^2y+2xy^2+y^3)\geq0.$$ The following inequality is also true.

Let $x$, $y$ and $z$ be non-negative numbers. Prove that: $$(x^5+y^5+z^5)^2\geq(x^2+y^2+z^2)(x^7y+y^7z+z^7x).$$