TITLE: An $\arcsin$ inequality
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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!