[不等式] Prove that $(\ln{\dfrac{1-\sin{xy}}{1+\sin{xy}}})^2 \geqslant \cdots$
Prove that if $x, y \in (0,\sqrt{\frac{\pi}{2}})$ and $ x \neq y$, then $(\ln{\dfrac{1-\sin{xy}}{1+\sin{xy}}})^2 \geqslant \ln{\dfrac{1-\sin{x^2}}{1+\sin{x^2}}}\ln{\dfrac{1-\sin{y^2}}{1+\sin{y^2}}}$