I conjecture this inequality $\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$ Let $x,y,z>0$,prove or disprove
$$\sqrt[4]{\dfrac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\dfrac{(x+y)(y+z)(z+x)}{8}}$$
I tried many times ,use
$$x^2+y^2+z^2\ge\dfrac{1}{3}(x+y+z)^2$$
and kown $$9(x+y)(y+z)(z+x)\ge 8(x+y+z)(xy+yz+xz)$$

Let $x+y+z=3u$, $y+xz+yz=3v^2$ and $xyz=w^3$.
Hence, it's obvious that our inequality is equivalent to $f(w^3)\geq0$, where $f(w^3)=w^3+A(u,v^2)$, 
which says that it's enough to prove our inequality for the minimal value of $w^3$.
But $x$, $y$ and $z$ are positive roots of the equation
$$(X-x)(X-y)(X-z)=0$$ or
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$w^3=X^3-3uX^2+3v^2X$$
which says that a line $Y=w^3$ and a graph of $Y=X^3-3uX^2+3v^2X$ have three common points.
Thus, $w^3$ gets a minimal value, when a line $Y=w^3$ is a tangent line to the graph of $Y=X^3-3uX^2+3v^2X$, which happens for equality case of two variables.
Also we need to check the case $w^3\rightarrow0^+$. 
Two these cases give a trivial inequalities:


*

*For $y=z=1$ we get $$(x-1)^2(2048x^7+6439x^6+18822x^5+29481x^4+33108x^3+24105x^2+9094x+1319)\geq0$$

*For $y=1$ and $z\rightarrow0^+$ we get
$$(4096x^6-729x^5+9372x^4-4374x^3+9372x^2-729x+4096)x^3\geq0$$
Done!

