$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$ if $x^2+y^2=1$. Show that $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$. It is given that $x^2+y^2=1$. $x,y$ are positive real numbers. 
[From a Regional Mathematical Olympiad, 2013 in India]

$$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}\ge \frac{4}{1+x^2+1+y^2} + \frac{1}{1+xy}$$
Use of AM-HM Inequality for $n=2$
$$=\frac{4}{3}+\frac{1}{1+xy} \ge \frac{1}{\frac{5}{4}+\frac{xy}{2}}$$
Expanding out
$$\frac{7+4xy}{3(1+xy)} \ge \frac{4}{5+2xy}$$
Common Denominators
$$(7+4xy)(5+2xy) \ge 12(1+xy)$$
Justified through positive terms
$$(7+4u)(5+2u) \ge 12(1+u)$$
$$35+34u+8u^2 \ge 12+12u$$
$$8u^2 + 22u + 23 \ge 0$$
But the global minimum of the LHS is $\frac{63}{8}$ for all real values of $u$. So the inequality is strictly true for all valid $x$ and $y$.
