Prove the inequality $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}}$ when $x^2+y^2=1$ I have to prove the inequality
$$
\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}}
$$
when $x^2+y^2=1$, using Cauchy-Schwarz Inequality.
The RHS is equal to $\frac{12}{5+2xy}$.
I can prove, using C-S, that the RHS is $\geq \frac{9}{4+xy}$ or  $\geq \frac{12}{5+4xy}$ but I can't go further.

I can prove the inequality only using A.M.-G.M. inequality proving that $xy\leq\frac{1}{2}$ and simplifying the expression all together, but this is not what I want.  
Thanks 

For $xy\geq0$ by C-S $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}=\frac{3}{2+x^2y^2}+\frac{1}{1+xy}=$$
$$=\frac{4}{\frac{4(2+x^2y^2)}{3}}+\frac{1}{1+xy}\geq\frac{(2+1)^2}{\frac{4(2+x^2y^2)}{3}+1+xy}.$$
Id est, it's enough to prove that
$$\frac{9}{\frac{4(2+x^2y^2)}{3}+1+xy}\geq\frac{12}{5+2xy}$$ or
$$(1-2xy)(1+8xy)\geq0,$$ which is true by C-S again:
$$2=(x^2+y^2)(1^2+1^2)\geq(x+y)^2=1+2xy,$$ which gives $1-2xy\geq0.$
For $xy\leq0$ by C-S again we obtain
$$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}=\frac{1}{\frac{2+x^2y^2}{3}}+\frac{1}{1+xy}\geq\frac{4}{\frac{2+x^2y^2}{3}+1+xy}.$$
Thus, it's enough to prove that
$$\frac{4}{\frac{2+x^2y^2}{3}+1+xy}\geq\frac{12}{5+2xy}$$ or
$$x^2y^2+xy-2\leq0$$ or
$$(1-xy)(2+xy)\geq0,$$ which is obvious.
