# Prove that $| xy-\sqrt{(1-x^2)(1-y^2)}|\leq1$ where $|x|\leq1$ and $|y|\leq1$ [duplicate]

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?

## marked as duplicate by Carl Mummert, lab bhattacharjee trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 21 '18 at 14:06

• I'd write $x=\cos\alpha$ and $y=\cos\beta$ myself, and assume $\alpha$ and $\beta$ are between $0$ and $\pi$. – Lord Shark the Unknown Aug 17 '18 at 19:26
• Consider the 2 cases $\leq \pi/2$ and $\geq \pi/2$ and use $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$. – Diger Aug 17 '18 at 19:29
• Incidentally, $xy\pm\sqrt{(1-x^2)(1-y^2)}$ are the bounds on $\operatorname{Corr}(A,\,C)$ given $\operatorname{Corr}(A,\,B)=x,\,\operatorname{Corr}(B,\,C)=y$. – J.G. Aug 17 '18 at 19:38

Let $x=\cos\alpha$ and $y=\cos\beta$, where $\{\alpha,\beta\}\subset[0,\pi].$
Thus, $\sin\alpha\geq0$, $\sin\beta\geq0$ and $$\left|xy-\sqrt{(1-x^2)(1-y^2)} \right| = \left| \cos(\alpha + \beta) \right| \leq 1.$$
\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 $y$.