Berkeley Math Circle

Solution:
First we use the inequality $a + b \geq 2 \sqrt{a b}$ and get
$$\frac{1}{1-x^{2}} + \frac{1}{1-y^{2}} \geq \frac{2}{\sqrt{(1-x^{2})(1-y^{2})}}$$
Now we notice that
$$(1-x^{2})(1-y^{2}) = 1 + x^{2} y^{2} - x^{2} - y^{2} \leq 1 + x^{2} y^{2} - 2 x y = (1-x y)^{2}$$
which implies that
$$\frac{2}{\sqrt{(1-x^{2})(1-y^{2})}} \geq \frac{2}{1-x y}$$
and this completes the proof.