This inequality is wrong – see the accepted answer (it appears there is no general inequality for these two expressions).
On the left we have harmonic mean of pairwise geometric means, which obeys:
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$
On the right we have arithmetic means of pairwise harmonic means, obeying the same inequality (the first inequality here may not be true as well, waiting for proof in another question):
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$
According to Wolfram Alpha the following is true:
$$\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$$
But I have not been able to prove it so far.
It may help to transform the RHS in the following way:
$$\frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)=\frac{2}{3} \frac{xyz(x+y+z)+(xy+yz+zx)^2}{(x+y+z)(xy+yz+zx)-xyz}$$
Best Answer
Your inequality is wrong! For $y=z=1$ and $x\rightarrow+\infty$ we obtain:
$3\leq\frac{2}{3}\left(1+\frac{1}{2}+1\right)$, which is wrong.