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.
On
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: $$\sum_{cyc}\frac{(x-y)^2}{xy}\geq\frac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$ $$=\frac{4(x-y)^2}{xy+xz+yz}\geq\frac{4(x-y)^2}{z^2+xy+xz+yz}=\frac{4(x-y)^2}{(y+z)(x+z)}.$$
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.