I was working with the inequality,
$$\left|\sin^{-1}(x) + \sin^{-1}(y)\right|\le\frac{\pi}{2} , \{|x|,|y|\le1\}.$$
I was trying to find an inequality only in terms of x and y without containing any trigonometric functions. Then after some hit and trial I found following beautiful inequality,
$$\biggl|x|x| + y|y| \biggr|\le 1 , \{|x|,|y|\le1\}.$$
But the problem is I couldn't derive this inequality and couldn't show their equivalence. I know they are equivalent by plotting them in Desmos(Function Plotting Program). Can you derive the proof showing the equivalence of these two inequalities ?
Note : The above inequality is the condition for the following relation to satisy,
$$\sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$$
I had thought this formula works for every value of $|x|,|y|\le1$ but turns out it doesn't. It only works when x and y satisfy above inequality.
Best Answer
We need $$\cos(\sin^{-1}x+\sin^{-1}y)\ge0$$
$$\implies\sqrt{(1-x^2)(1-y^2)}\ge xy$$
which is obvious $xy<0$
What if $xy=0?$
If $xy>0,$ $$(1-x^2)(1-y^2)\ge x^2y^2\iff 1\ge x^2+y^2$$