Is $8(x^3+y^3+z^3)^2\ge 9(x^2+yz)(y^2+xz)(z^2+xy)$ true for nonnegative numbers?

Question

For $x,y,z\ge 0$ prove:$$8(x^3+y^3+z^3)^2\ge 9(x^2+yz)(y^2+xz)(z^2+xy)$$My attempt:

we shall prove that:$$8x^6+8y^6+8z^6+16x^3y^3+16x^3z^3+16y^3z^3\ge\\18x^2y^2z^2+9x^3y^3+9x^3z^3+9y^3z^3+9x^4yz+9y^4xz+9z^4xy$$here after I'm stuck. Can we keep on this idea or should try another one?

Thanks in advance!

Answer 1

The inequality is equivalant to \begin{eqnarray*} 8(x^6+y^6+z^6)+7(x^3y^3+y^3z^3+z^3x^3)-9(xyz^4+yzx^4+zxy^4)-18x^2y^2z^2 \geq 0 \end{eqnarray*} This is the same as \begin{eqnarray*} 8(x^2(x^2-yz)^2+y^2(y^2-zx)^2+z^2(z^2-xy)^2)+7(xy(xy-z^2)^2+yz(yz-x^2)^2+zx(zx-y^2)^2) \geq 0 \end{eqnarray*} which is clearly true.

Edit:

Answer 2

It is worth proceeding your way... for example like this:

$$8x^6+8y^6+8z^6+7x^3y^3+7y^3z^3+7x^3z^3-18x^2y^2z^2-9x^4yz-9y^4xz-9z^4xy\ge 0$$ $$4.5x^6-9x^4yz+4.5x^2y^2z^2+4.5y^6-9y^4xz+4.5x^2y^2z^2+4.5z^6-9z^4xy+4.5x^2y^2z^2+3.5(x^6+y^6+z^6)-31.5x^2y^2z^2\ge 0$$

$$4.5(x^2(x^2-yz)^2+y^2(y^2-zx)^2+z^2(z^2-xy)^2)+3.5(x^6+y^6+z^6-3x^2y^2z^2)+7(x^3y^3+y^3z^3+x^3z^3-3x^2y^2z^2)\ge 0$$

The first term is obviously $\ge 0$, while the second and third term are $\ge 0$ because of the (known, AM/GM) inequality $a^3+b^3+c^3\ge 3abc$, applied to $x^2, y^2,z^2$ and $xy, yz, xz$, respectively.

Answer 3

The hint:

Use the following Schur: $$4.5\sum_{cyc}(x^6-x^4y^2-x^4z^2+x^2y^2z^2)\geq0.$$

Answer 4

All can be solved by Weighted AM-GM. We will continue from the expansion $8 \sum x^6 + 7\sum x^3y^3 \geq 18x^2y^2z^2 + 9\sum x^4yz$, which is the one you gave us.

We know that $6 \sum x^6 + 3 \sum y^3z^3 \geq 9 \sum x^4yz$.

And also $2 \sum x^6 + 4\sum y^3z^3 \geq 18x^2y^2z^2$