# 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!

• @Micheal Rozenberg ... are $x$,$y$ and $z$ non negative ? ... I have a proof, but it relies on this assumption. Mar 18, 2017 at 22:13
• @Donald Splutterwit The variables are reals. Mar 19, 2017 at 3:40
• I don't have your answer, but note that the equation after your second "or" has an error: The RHS is degree 4. Did you mean $\frac{20}{3} u^3(27u^2-18uv^2+w^3)$? Aug 30, 2017 at 19:22
• @Mark Fischler It was typo. It should be $\frac{2}{9}(27u^2-18uv^2+w^3)\prod\limits_{cyc}(x-y)$. I fixed. Thank you! Aug 30, 2017 at 20:11
• Note that $\frac{(x^2+y^2+z^2+xy+yz+zx)^3}{8} \ge \frac{(x+y+z)^6}{27}$ for all real numbers. This inequality is stronger than math.stackexchange.com/questions/2063091/… Sep 7, 2019 at 6:38

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 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 & 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 & 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*}
• I have a proof for non-negative variables. I am looking a proof for all real variables. For non-negative variables even the following is true: $\sum\limits_{cyc}(4x^6+5x^5y)\geq(x^2+y^2+z^2)^3$. Mar 19, 2017 at 4:33