contest math - Help proving this inequality

Give $x$ , $y$ , with $x,y \in \Re$ , and $x^2 + y^2 = 1$ , show that .
$$ \left(\frac{1}{1+x^2}\right) + \left(\frac{1}{1+y^2}\right) + \left(\frac{1}{1+xy}\right) \ge \left(\frac{3}{1+((x+y)/2))^2}\right)$$

This is my approach:

from $x^2 + y^2 = 1$ , we can see that $xy \le \frac{1}{2}$ , which says $1+xy$ is always positive
Since we know $x^2 \ge 0$ and $y^2 \ge 0$ , we can say that $x^2 + 1 $ and $y^2 +1 $ , the denominators of the expression are always positive .

Now evaluating $\left(\frac{3}{1 + \left(\frac{x+y}{2}^2\right) }\right)$ , we can see it has a lower bound of 2 , following from $xy \le \frac{1}{2}$ .
Thus we are left to show that $$ \left(\frac{1}{1+x^2}\right) + \left(\frac{1}{1+y^2}\right) + \left(\frac{1}{1+xy}\right) \ge 2$$

Which follows directly from Titu's lemma .

Now this is wrong , since this concludes nothing .Can anyone give me some hints on which way to proceed ?

Answer

Let $xy=t$.

Thus, by AM-GM $\frac{1}{2}\geq t\geq-\frac{1}{2}$ and we need to prove that $$\frac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}+\frac{1}{1+xy}\geq\frac{3}{1+\frac{x^2+y^2+2xy}{4}}$$ or

$$\frac{3}{2+t^2}+\frac{1}{1+t}\geq\frac{3}{1+\frac{1+2t}{4}}$$ or

$$(1-2t)(5t^2+3t+1)\geq0$$ and we are done!