$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$ 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]
 A: $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}\ge \frac{4}{1+x^2+1+y^2} + \frac{1}{1+xy}$$
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 $y$.
A: $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2} \rightarrow \frac{1+x^2+1+y^2}{(1+x^2)(1+y^2)}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2}  \rightarrow \frac{3}{2+x^2y^2}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2}  \rightarrow \frac{3+3xy+2+2x^2y^2}{2+2xy+x^2y^2+x^3y^3}\ge \frac{12}{4+x^2+y^2+2xy}  \rightarrow \frac{5+3xy+2x^2y^2}{2+2xy+x^2y^2+x^3y^3}\ge \frac{12}{5+2xy}.$$ 
 Consider $xy=z$ then we have:
$$\frac{5+3z+2z^2}{2+2z+z^2+z^3}\ge \frac{12}{5+2z}. $$ We know that $$x^2+y^2 \ge 2 (\sqrt{x^2y^2})\rightarrow -0.5\le z \le 0.5 \rightarrow (5+2z)\ge 0, (z+1)\ge 0, (1+2z) \ge 0$$. So we have:
$$\frac{5+3z+2z^2}{2+2z+z^2+z^3} - \frac{12}{5+2z}\ge 0 \rightarrow \frac{25+25z+16z^2+4z^3-24-24z-12z^2-12z^3}{(2+2z+z^2+z^3)(5+2z)} \rightarrow \frac{1+z+4z^2-8z^3}{(z+1)(2+z^2)(5+2z)}\ge 0 \rightarrow 1+z+4z^2-8z^3 \ge 0 \rightarrow (1+z) + 4z^2(1-2z) \ge 0$$
