Let $|x|\leq1, |y|\leq 1$. Show that$0 \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy \sqrt{1-x^{2}} \sqrt{1-y^{2}} \leq 1$ I would appreciate if somebody could help me with the following problem
Let $|x|\leq1, |y|\leq 1$. Show that
$$
0  \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy  \sqrt{1-x^{2}}  \sqrt{1-y^{2}} \leq 1
$$

Use the fact that for $|x|\le 1,|y|\le 1$, $$\arccos x +\arccos y =\arccos \left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)  $$ or $$\cos \left(\arccos x +\arccos y \right) = \left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)  $$ and hence we can say 
$$-1 \le \left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right) \le 1$$
Then the given expression becomes
$$0 \leq (x+y)^2 \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy  \sqrt{1-x^{2}}  \sqrt{1-y^{2}} \leq (x-y)^2 \leq 1$$
