Prove that $\sum\limits_{cyc}(4x^6+5x^5y)\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^3}{8}$ 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!
6*(8*(4*(x^6+y^6+z^6)+5*(x^5*y+y^5*z+z^5*x))-(x^2+y^2+z^2+x*y+y*z+z*x)^3)-(75*((x-y)^6+(x-z)^6+(z-y)^6)+432*(y*z*(z-y)^4+y*x*(x-y)^4+x*z*(z-x)^4)+567*(y^2*z^2*(z-y)^2+y^2*x^2*(x-y)^2+x^2*z^2*(z-x)^2)+36*(x^2(x^2-y*z)^2+y^2*(y^2-x*z)^2+z^2*(z^2-y*x)^2)+120*(x*y*(x^2-y*z)^2+y*z*(y^2-x*z)^2+z*x*(z^2-y*x)^2)+120*(y^3*z*(y-x)^2+z^3*x*(y-z)^2+y*x^3*(x-z)^2)+89*x*y*z*(x*(x-y)^2+y*(y-z)^2+z*(z-x)^2)+169*x*y*z*(x*(x-z)^2+y*(y-x)^2+z*(z-y)^2)+39*x*y*z*(x*(y-z)^2+y*(x-z)^2+z*(x-y)^2));


The above can be copied &amp; pasted into Reduce (Computer Algebra package). It should evaluate to zero.

The following expression is clearly non-negative ...
\begin{eqnarray*}
75\sum_{cyc}(x-y)^6+432\sum_{cyc}xy(x-y)^4+567\sum_{cyc}x^2y^2(x-y)^2+36\sum_{cyc}x^2(x^2-yz)^2+120\sum_{cyc}xy(x^2-yz)^2+120\sum_{cyc}x^3y(x-y)^2+89xyz\sum_{cyc}x(x-y)^2+169xyz\sum_{cyc}y(x-y)^2+39xyz\sum_{cyc}x(y-z)^2 \geq 0
\end{eqnarray*}
Expanding these out &amp; removing the factor of $6$ ... we have 
\begin{eqnarray*}
8\sum_{cyc} (4 x^6+5x^5y) \geq \sum_{cyc} x^6 +3\sum_{perms} x^5y +6\sum_{perms} x^4y^2 +7\sum_{cyc} x^3y^3 +9\sum_{perms} x^4yz +15\sum_{perms} x^3y^2z +21 x^2y^2z^2 
\end{eqnarray*}
Now note that the RHS is a perfect cube &amp; thus we have the required result
\begin{eqnarray*}
\sum_{cyc} (4 x^6+5x^5y) \geq \frac{1}{8} \left(\sum_{perms} x^2+yz \right)^3 \end{eqnarray*}
6*(8*(4*(x^6+y^6+z^6)+5*(x^5*y+y^5*z+z^5*x))-(x^2+y^2+z^2+x*y+y*z+z*x)^3)-(75*((x-y)^6+(x-z)^6+(z-y)^6)+432*(y*z*(z-y)^4+y*x*(x-y)^4+x*z*(z-x)^4)+567*(y^2*z^2*(z-y)^2+y^2*x^2*(x-y)^2+x^2*z^2*(z-x)^2)+36*(x^2(x^2-y*z)^2+y^2*(y^2-x*z)^2+z^2*(z^2-y*x)^2)+120*(x*y*(x^2-y*z)^2+y*z*(y^2-x*z)^2+z*x*(z^2-y*x)^2)+120*(y^3*z*(y-x)^2+z^3*x*(y-z)^2+y*x^3*(x-z)^2)+89*x*y*z*(x*(x-y)^2+y*(y-z)^2+z*(z-x)^2)+169*x*y*z*(x*(x-z)^2+y*(y-x)^2+z*(z-y)^2)+39*x*y*z*(x*(y-z)^2+y*(x-z)^2+z*(x-y)^2));


The above can be copied &amp; pasted into Reduce (Computer Algebra package). It should evaluate to zero.

The following expression is clearly non-negative ...
\begin{eqnarray*}
75\sum_{cyc}(x-y)^6+432\sum_{cyc}xy(x-y)^4+567\sum_{cyc}x^2y^2(x-y)^2+36\sum_{cyc}x^2(x^2-yz)^2+120\sum_{cyc}xy(x^2-yz)^2+120\sum_{cyc}x^3y(x-y)^2+89xyz\sum_{cyc}x(x-y)^2+169xyz\sum_{cyc}y(x-y)^2+39xyz\sum_{cyc}x(y-z)^2 \geq 0
\end{eqnarray*}
Expanding these out &amp; removing the factor of $6$ ... we have 
\begin{eqnarray*}
8\sum_{cyc} (4 x^6+5x^5y) \geq \sum_{cyc} x^6 +3\sum_{perms} x^5y +6\sum_{perms} x^4y^2 +7\sum_{cyc} x^3y^3 +9\sum_{perms} x^4yz +15\sum_{perms} x^3y^2z +21 x^2y^2z^2 
\end{eqnarray*}
Now note that the RHS is a perfect cube &amp; thus we have the required result
\begin{eqnarray*}
\sum_{cyc} (4 x^6+5x^5y) \geq \frac{1}{8} \left(\sum_{perms} x^2+yz \right)^3 \end{eqnarray*}