Prove this Complicated Inequality

$x = a+b,y= b+c,z = c+a$, then

$$x^2 + y^2 + z^2 \le 4$$

problem becomes

$$\sum\frac{4 + (x+z-y)(x+y-z)}{4x^2}=\sum \frac{4 + x^2 - y^2 - z^2 + 2yz}{4x^2}$$

$$=\sum\frac{4-x^2 - y^2 -z^2 + 2x^2 + 2yz}{4x^2}\ge \sum\frac{2x^2 + 2yz}{4x^2} $$

$$=\sum \left(\frac{1}{2} + \frac{1}{2}\frac{yz}{x^2}\right)\ge \frac{3}{2} +\frac{1}{2}\cdot 3\left(\prod\frac{yz }{x^2}\right)^{1/3} = 3$$