Because $$\frac{(x-1)^2}{y}+\frac{(y-1)^2}{z}\geq\frac{(x+y-2)^2}{y+z}=\frac{z^2}{y+z}.$$

By T2 lemma we can write : $$\sum\frac{(x-1)^2}{y} \geq \frac{1}{2}\sum \frac{(x+y-2)^2}{y+z} = \frac{1}{2}\sum \frac{z^2}{y+z}$$Now we need prove : $2\sum \frac{z^2}{y+z} \geq \frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}$ Then .......?! $\blacksquare$

@above well $2z^2 - (y^2 + z^2) = z^2 - y^2$, so you want $\sum \frac{z^2-y^2}{y+z} = \sum (z-y) = 0$ to be at least $0$, done.

This can be generalized for $x,y,z\in \mathbf{R^+}$ such that $x+y+z=2k$ $$\sum_{cyc}{\frac{(x-k)^2}{y}}\geq \frac{1}{4}\left(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}\right)$$

Arqady's solution below destroy it in a minute.