BĐT cần chứng minh tương đương với:
$\left|\dfrac{(x^2-y^2)(1-x^2y^2)}{(1+x^2)^2(1+y^2)^2}\right|\le\dfrac{1}{4} (*)$
Vì: $\left(\dfrac{x-y}{\sqrt{(1+x^2)(1+y^2)}}\right)^2+\left(\dfrac{1+xy}{\sqrt{(1+x^2)(1+y^2)}}\right)^2=1$
và $\left(\dfrac{x+y}{\sqrt{(1+x^2)(1+y^2)}}\right)^2+\left(\dfrac{1-xy}{\sqrt{(1+x^2)(1+y^2)}}\right)^2=1$.
Đặt $\cos\alpha=\dfrac{x-y}{\sqrt{(1+x^2)(1+y^2)}};\sin\alpha=\dfrac{1+xy}{\sqrt{(1+x^2)(1+y^2)}}$
$\cos\beta=\dfrac{x+y}{\sqrt{(1+x^2)(1+y^2)}};\sin\beta=\dfrac{1-xy}{\sqrt{(1+x^2)(1+y^2)}}$
Khi đó:
$(*) \Leftrightarrow |\cos\alpha\sin\beta\cos\beta\sin\alpha|\leq\dfrac{1}{4} \Leftrightarrow |\dfrac{1}{4}\sin2\alpha\sin2\beta|\leq\dfrac{1}{4}$, luôn đúng.