<s>[INST] Prove $\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \geq \sqrt{\frac{xy+yz+zx}{3}}$ for $x,y,z \geq 0$ We have geometric mean of pairwise arithmetic means on the left, which obeys the following inequality:
$$\frac{x+y+z}{3}  \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq \sqrt[3]{xyz}$$
And on the right we have root-mean-square of geometric means, obeying the same inequality:
$$\frac{x+y+z}{3}  \geq \color{blue}{\sqrt{\frac{xy+yz+zx}{3}} } \geq \sqrt[3]{xyz}$$
This time I checked with Wolfram Alpha first, and apparently, the inequality in the title is true for all $x,y,z \geq 0$:
$$\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \geq \sqrt{\frac{xy+yz+zx}{3}}$$


How do we prove this inequality for  for all $x,y,z \geq 0$?

We need to prove:
$$27(x+y)^2(y+z)^2(z+x)^2 \geq 64(xy+yz+zx)^3$$
Expanding directly leads to a very complicated expression, and overall this inequality is a little tight. I have not been able to prove it by AM-GM or Cauchy.
 [/INST] Below is a simple and nice proof of
$$27(x+y)^2(y+z)^2(z+x)^2 \geq 64(xy+yz+zx)^3 \quad \forall x,y,z \ge 0.$$
Indeed, it is straightforward to see that the above inequality follows directly from
\begin{equation}
9(x+y)(y+z)(z+x) \ge 8(x+y+z)(xy+yz+zx) \quad (1)
\end{equation}
and 
\begin{equation}
(x+y+z)^2 \ge 3(xy+yz+zx) \quad (2)
\end{equation}
To prove $(1)$, just notice that 
$$LHS - RHS = \sum x(y^2+z^2) - 6xyz = \sum x(y-z)^2 \ge 0.$$
Remark. From the proof, we have 
$$\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \geq \sqrt[3]{\frac{(x+y+z)(xy+yz+zx)}{9}} \geq \sqrt{\frac{xy+yz+zx}{3}}.$$
</s>