Show that if $0<|x|,|y|<1$, then $$\arcsin |x| +\arcsin |y| > \arcsin\left|\frac{x+y}{1+xy}\right|.$$
I found a proof (see below). Is there a different way (hopefully simpler) to show that the above inequality holds? Any reference to similar inequalities?
Proof. Let $a=\mathrm{arctanh}(x)\not=0$, $b=\mathrm{arctanh}(y)\not=0$ and let $$h(t)=\arcsin(\tanh(t)).$$ Then the inequality is equivalent to $$h(|a|)+h(|b|)>h(|a+b|).$$ Now, $h(0)=0$, $h$ is monotone and strictly concave in $(0,+\infty)$ (note that $h''(t)=-\sqrt{1-\tanh^2(t)}\tanh(t)$). Hence $$h(|a|)>\frac{|b|h(0)}{|a|+|b|}+\frac{|a|h(|a|+|b|)}{|a|+|b|}=\frac{|a|h(|a|+|b|)}{|a|+|b|}.$$ Similarly $$h(|b|)>\frac{|b|h(|a|+|b|)}{|a|+|b|},$$ After adding together the last two inequalities we have $$h(|a|)+h(|b|)>h(|a|+|b|)\geq h(|a+b|)$$ and we are done.
It's enough to prove that $$\cos\left(\arcsin|x|+\arcsin|y|\right)<\cos\arcsin\left|\frac{x+y}{1+xy}\right|$$ or $$\sqrt{(1-x^2)(1-y^2)}-|xy|<\sqrt{1-\left(\frac{x+y}{1+xy}\right)^2}$$ or $$|xy|>\sqrt{(1-x^2)(1-y^2)}-\frac{\sqrt{(1-x^2)(1-y^2)}}{1+xy}$$ or $$|xy|(1+xy)>xy\sqrt{(1-x^2)(1-y^2)},$$ for which it's enough to prove that $$(1+xy)^2>(1-x^2)(1-y^2)$$ or $$(x+y)^2>0.$$ The equality does not occur because for $x+y=0$ we obtain $xy<0$.
Done!