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?

We have
\begin{align*}
 \sum_{\mathrm{cyc}} \frac{xy + 1}{(x + y)^2}
 &\ge \sum_{\mathrm{cyc}} \frac{xy + \frac{x^2 + y^2 + z^2 + (x + y + z)^2}{4}}{(x + y)^2}\\
 &= \sum_{\mathrm{cyc}} \frac{2xy + x^2 + y^2 + z^2 + xy + yz + zx}{2(x + y)^2}\\
 &= \sum_{\mathrm{cyc}} \frac{(x + y)^2 + (y + z)(z + x)}{2(x + y)^2}\\
 &= \frac{3}{2}  + \sum_{\mathrm{cyc}} \frac{(y + z)(z + x)}{2(x + y)^2}\\
 &\ge \frac32 + 3\sqrt[3]{\prod_{\mathrm{cyc}}\frac{(y + z)(z + x)}{2(x + y)^2}}\\
 &= 3.
\end{align*}
We are done.
