<s>[INST] Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$ Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$
I tried:
$x=\sin\alpha$ and $y=\cos\beta$
$\sqrt{(1-x^2)(1-y^2)}=\sqrt{\cos^2\alpha\sin^2\beta}$ but if I write $\sqrt{\cos^2\alpha\sin^2\beta}=\cos\alpha \sin\beta$, it's not true because 
$\cos\alpha \sin\beta$ can be negative. 
Can someone give me an idea?
 [/INST] $$\begin{align*}xy-\sqrt{(1-x^2)(1-y^2)}&\le 1\\
\Leftrightarrow
(xy)^2&\le 1+(1-x^2-y^2+(xy)^2 )+2\sqrt{1-x^2)(1-y^2)}
\end{align*}$$
which is true since $|x|\le 1$ then $1-x^2 > 0$ and similarly for $y$.
And do the same for the other side.
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