How to prove $\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}$ for $x,y,z>0$? 
Prove that for $x,y,z$ positive numbers:
  $$
\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}
$$

I tried to apply MA-MG inequality:
$x+y≥2\sqrt{xy}$ and the others and multiply them but it becomes
$\frac{(x+y)(y+z)(z+x)}{4xyz}≥2$ but $\frac{x+z}{y+z}+\frac{y+z}{x+z}≥2$ so it doesn't work like that.
I don't know what to apply here. 

We have
$$
\frac1 {4yz}+\frac1 {4xz}\ge \frac1{(y+z)^2}+\frac1{(x+z)^2}.
$$ Therefore
$$
\frac{(x+y)}{4xyz}(y+z)(z+x)\ge\left(\frac1{(y+z)^2}+\frac1{(x+z)^2}\right)(y+z)(z+x),
$$ which leads to the desired inequality.
