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.
 A: First we prove a very close inequality:

$$27(x+y)^2(y+z)^2(z+x)^2 \geq 63(xy+yz+zx)^3+27x^2y^2z^2$$

The proof is as follows. By direct expansion we see:
$$(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)-xyz$$

$$(x+y)^2(y+z)^2(z+x)^2= \\ =(x+y+z)^2(xy+yz+zx)^2-2xyz(x+y+z)(xy+yz+zx)+x^2y^2z^2$$
Using Cauchy-Schwartz we can transform the second term:
$$(x+y+z)(xy+yz+zx) \geq 9xyz$$

$$(x+y)^2(y+z)^2(z+x)^2 \geq \\ \geq (x+y+z)^2(xy+yz+zx)^2-\frac{2}{9}(x+y+z)^2(xy+yz+zx)^2+x^2y^2z^2$$

$$(x+y)^2(y+z)^2(z+x)^2 \geq \frac{7}{9}(x+y+z)^2(xy+yz+zx)^2+x^2y^2z^2= \\ = \frac{7}{9}(x^2+y^2+z^2+2(xy+yz+xz))(xy+yz+zx)^2+x^2y^2z^2 \geq \\ \geq \frac{7}{3}(xy+yz+zx)^3+x^2y^2z^2$$
Finally we get:
$$3(x+y)^2(y+z)^2(z+x)^2 \geq 7(xy+yz+zx)^3+3x^2y^2z^2$$

Now we prove the original inequality.
First, let's get back and rearrange the original equality we used:
$$(x+y)(y+z)(z+x)+xyz=(x+y+z)(xy+yz+zx)$$
Now let's square this and denote:
$$A=(x+y)(y+z)(z+x)$$ 
$$B=(xy+yz+zx)$$

$$A^2+2xyzA+x^2y^2z^2=(x+y+z)^2B^2$$
$$A^2=(x+y+z)^2B^2-2xyzA-x^2y^2z^2 \geq 3B^3-\frac{2}{\sqrt{27}} \sqrt{B^3} A-A^2+\frac{7}{3}B^3$$

$$A^2+\frac{1}{\sqrt{27}} \sqrt{B^3} A-\frac{8}{3}B^3 \geq 0 \tag{1}$$

Here we used the inequality from the previous section and two well known inequalities:
$$x^2y^2z^2 \leq A^2-\frac{7}{3}B^3$$
$$27x^2y^2z^2 \leq B^3$$
$$(x+y+z)^2 \geq 3B$$
$(1)$ is a quadratic inequality, and is easily solved:

$$A \geq \frac{8}{\sqrt{27}} \sqrt{B^3}$$

The original inequality obviously follows.
A: 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 
$$\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \geq \sqrt[3]{\frac{(x+y+z)(xy+yz+zx)}{9}} \geq \sqrt{\frac{xy+yz+zx}{3}}.$$
