A rational inequality involving three pairwise sums

Show that $\forall x,y,z>0$ we have: $\frac{(x+y)(y+z)(z+x)}{4xyz} \geq \frac{x+z}{y+z} + \frac{y+z}{x+z}$ .

made by Caragea C.

Here is what I have found:

$\frac{x+y}{4xyz}= \frac{1}{4yz} +\frac{1}{4xz} \geq \frac{1}{(y+z)^2}+ \frac{1}{(x+z)^2}$ and so we have that: $\frac{x+y}{4xyz} \geq \frac{1}{(y+z)^2} + \frac{1}{(x+z)^2}$ and we multiply by $(y+z)(x+z)$ and we have our ineq!

do you have other solutions? cheers!

Source: InternLM Lean-Workbook, record lean_workbook_127 (Apache-2.0). Complete source proposition preserved; proof developed independently.