Let $x$, $y$ and $z$ be real numbers. Prove that: $$4(x^6+y^6+z^6)+5(x^5y+y^5z+z^5x)\geq\frac{(x^2+y^2+z^2+xy+xz+yz)^3}{8}$$
I think cyclic homogeneous polynomial sixth degree inequalities with three variables are interesting enough because it's still open.
By the way, the case of fifth degree and less is very easy and we can kill it by $uvw$.
My trying by $uvw$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $$\sum_{cyc}(32x^6+40x^5y)\geq(9u^2-3v^2)^3$$ or $$\sum_{cyc}(32x^6+20x^5y+20x^5z)-27(3u^2-v^2)^3\geq20\sum_{cyc}(x^5z-x^5y)$$ or $$4w^6+516u^3w^3-244uv^2w^3+2592u^6-4644u^4v^2+1872u^2v^4-72v^6-3(3u^2-v^2)^3\geq$$ $$\geq\frac{20}{9}(x-y)(y-z)(z-x)(27u^3-18uv^2+w^3).$$ We can show that the left side is non-negative and after squaring of the both sides
we'll get a fourth degree inequality of $w^3$, which is nothing I think.
Any hint would be desirable.
Thank you!