Prove $\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq 3$ when $x^2+y^2+z^2+(x+y+z)^2\leq 4$. 
Let $x,y,z\in \Bbb{R}^+$ such that $x^2+y^2+z^2+(x+y+z)^2\leq 4$. Prove that $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq 3.$$

As there are three fractions in the left side and a single term in the right side, I thought Cauchy-Schwarz might be of help. But from Cauchy-Schwarz, I got $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq \frac{\left(\sum_{cyc}\sqrt{xy+1}\right)^2}{x^2+y^2+z^2+(x+y+z)^2}.$$
Then we have to prove that $$\left(\sum_{cyc}\sqrt{xy+1}\right)^2\geq12.$$
But this doesn't seem to work. So, how to solve the problem?

$$\sum_{cyc}\frac{xy+1}{(x+y)^2}-3\geq\sum_{cyc}\frac{2xy+\sum\limits_{cyc}(x^2+xy)}{2(x+y)^2}-3=\frac{\sum\limits_{cyc}(x^2+3xy)\sum\limits_{cyc}(x^4-x^2y^2)}{2\prod\limits_{cyc}(x+y)^2}\geq0.$$
