Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
Assuming  that :
$\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ 
So
$(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$
$\sqrt{(x-1)(y-1)} \leq xy-x-y+2$
$ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$
Here I'm stuck !

WLOG $\sqrt x=\sec A,\sqrt y=\sec B$ where $0<A,B<\dfrac\pi2$
$\sqrt{x-1}+\sqrt{y-1}=\tan A+\tan B=\dfrac{\sin(A+B)}{\cos A\cos B}\le\sqrt{\sec A\sec B}=?$
