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 need to prove that
$$\frac{(x+y)(x+z)(y+z)}{4xyz}-2\geq\frac{x+z}{y+z}+\frac{y+z}{x+z}-2$$ or
$$\sum_{cyc}\frac{(x-y)^2}{xy}\geq\frac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
$$\sum_{cyc}\frac{(x-y)^2}{xy}\geq\frac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
$$=\frac{4(x-y)^2}{xy+xz+yz}\geq\frac{4(x-y)^2}{z^2+xy+xz+yz}=\frac{4(x-y)^2}{(y+z)(x+z)}.$$
