How $2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}$ and $\frac{3xyz}{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}$ are related for $x,y,z \geq 0$? The first expression is a heometric mean of pairwise harmonic means and obeys the following inequality:
$$\sqrt[3]{xyz} \geq \color{blue}{ 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}} \geq  \frac{3xyz}{xy+yz+zx}$$ 
The second expression is a harmonic mean of pairwise heometric means and obeys the same inequality:
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{3xyz}{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}} \geq  \frac{3xyz}{xy+yz+zx}$$

I want to know how these two expressions are related ($\leq$ or $\geq$) for $x,y,z \geq 0$ .


Edit
Since the realtion is homogenous, we can take $xyz=1$ without loss of generality, then we just need to compare:
$$\frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} \text{ ? } \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$$
Or rather:
$$\frac{\sqrt[3]{(x+y)(y+1/(xy))(1/(xy)+x)}}{2} \text{ ? } \frac{\sqrt{x}+\sqrt{y}+\sqrt{1/(xy)}}{3}$$
 A: I will prove
$$\frac{3xyz}{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}\ge 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}$$
Let $x=a^2,y=b^2,z=c^2$. Then, after taking third powers, our inequality becomes
$$\frac{27a^6b^6c^6}{(a+b+c)^3a^3b^3c^3}\ge 8{\frac{a^4b^4c^4}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}$$
After expanding, one needs to show
$$27(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge 8(a+b+c)^3abc\qquad (1)$$
or 
$$27\sum_{sym}u^4v^2+54a^2b^2c^2\ge 4\sum_{sym}u^4vw+24\sum_{sym}u^3v^2w+48a^2b^2c^2$$
After canceling both sides, 
$$27\sum_{sym}u^4v^2+\sum_{sym}u^2v^2w^2\ge 4\sum_{sym}u^4vw+24\sum_{sym}u^3v^2w$$
By AM-GM, 
$$\sum_{sym}u^4v^2+\sum_{sym}u^2v^2w^2\ge \sum_{sym}u^3v^2w$$
And by Muirhead's inequality
$$\sum_{sym}u^4v^2\ge \sum_{sym}u^3v^2w$$
$$\sum_{sym}u^4v^2\ge \sum_{sym}u^4vw$$
Summing these all up gives us the desired result. Observe that $(1)$ can be proved in million different ways, I just like Muirhead's Inequality.
A: In Emre's solution, he transformed the problem into
$$27(a^2+b^2)(b^2+c^2)(c^2+a^2)\ge 8(a+b+c)^3abc\qquad (1)$$ 
and then proved this inequality by fully expanding it. 
In this answer, I will give a simpler proof. 
Lemma. For any $x,z,z\ge 0$ we have $$\begin{equation}
9(x+y)(y+z)(z+x) \ge 8(x+y+z)(xy+yz+zx) \quad (1)
\end{equation}$$
Proof. True because 
$$LHS - RHS = \sum x(y^2+z^2) - 6xyz = \sum x(y-z)^2 \ge 0.$$
This well-known lemma is very useful in solving olympiad inequalities. I have mentioned it once here.
Applying the lemma, $(1)$ is reduced to proving
$$24(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2) \ge 8(a+b+c)^3abc,$$
which follows directly from $3(a^2+b^2+c^2) \ge (a+b+c)^2$ and $a^2b^2+b^2c^2+c^2a^2 \ge abc(a+b+c)$ (the latter is $p^2+q^2+r^2 \ge pq+qr+rp$ with $p=ab,q=bc,r=ca$).
