Just set $ x= \cos\alpha $ , $y= \cos \beta $, $ \alpha,\beta \in [0;\pi] $ then the rest is quite easy

hello, the searched maximium is $2$ and will be attained for $(x,y)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$. Sonnhard.

WakeUp wrote: Find the largest value of the expression \[xy+x\sqrt{1-x^2}+y\sqrt{1-y^2}-\sqrt{(1-x^2)(1-y^2)}\] By C-S $ (xy-\sqrt{(1-x^2)(1-y^2)}+x \sqrt{1-x^2}+y\sqrt{1-y^2})^2 $ $ \le 2[ (xy-\sqrt{(1-x^2)(1-y^2)})^2+(x\sqrt{1-x^2}+y\sqrt{1-y^2})^2] $ $ \le 2 (2x^2y^2-x^4-y^4+1 ) \le 2 $

Sorry to revive, but Dr Sonnhard Graubner wrote: hello, the searched maximum is $2$ and will be attained for $(x,y)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$. Sonnhard. does not work, as the value would be 1. I got : Maximize $\frac{\sin 2a}{2} + \frac{\sin 2b}{2} - \cos (a+b)$ for $0\le a,b \le \frac{\pi}{2}$ which is equal to $\sin(a+b)\cos(a-b) - \cos(a+b)$ which has maximum magnitude $\sqrt{1^2 + \cos(a-b)^2}$. Thus, the maximum is $\sqrt{2}$.