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 $$9(x+y)(y+z)(z+x)\ge 8(x+y+z)(xy+yz+xz)$$

This inequality is nice. And without any advanced techniques/theorems, it is also hard.
The technique that Michael used in his solution was developed more than ten years ago (credit goes to Nguyen Anh Cuong who was, to the best of my knowledge, the first to introduce the idea, under the name "ABC method"). The general idea is the following.
Suppose that we need to prove $F(p,q,r)\ge 0$ where $p=x+y+z,q=xy+yz+zx,r=xyz$ and $x,y,z\ge 0$. If $f(r):= F(p,q,r)$ is a concave function (w.r.t. $r$) then it suffices to prove the inequality for $(x-y)(y-z)(z-x)=0$ or $xyz=0$.
Let me introduce another nice and "simple" (not quite!) solution. 
We will show that, for any $x,y,z\ge 0$:
$$\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge \sqrt[4]{\frac{(x+y)(y+z)(z+x)(x+y+z)}{24}} \ge \sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$$
Although these new inequalities are tighter, they are much easier to solve.
Indeed: 


*

*The left-hand side inequality is equivalent to
$$\boxed{8(xy+yz+xz)(x^2+y^2+z^2) \ge 3(x+y)(y+z)(z+x)(x+y+z)}$$
(boxed, because I believe this is an important lemma that Olympiad students should keep in mind). Denote $A=xy(x^2+y^2)+yz(y^2+z^2)+zx(z^2+x^2),B=x^2y^2+y^2z^2+z^2x^2$ and $C=xyz(x+y+z)$ then the above inequality becomes $$8(A+C) \ge 3(A+2B+4C),$$
which is true because $A\ge 2B\ge 2C$.

*The right-hand side inequality is equivalent to $$8(x+y+z)^3\ge 27(x+y)(y+z)(z+x),$$ which is just AM-GM: $(a+b+c)^3 \ge 27abc$ where $a=x+y,b=y+z,c=z+x$.


We are done.
