If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\pi-\arcsin z$
$x\sqrt{1-y^2}+y\sqrt{1-x^2}=z$
Similarly,$y\sqrt{1-z^2}+z\sqrt{1-y^2}=x$
Similarly,$x\sqrt{1-z^2}+z\sqrt{1-x^2}=y$
Adding the three equations,we get
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=\frac{x+y+z}{2}$
I am stuck here,please help me.Thanks.

Hint:
Write
$$x = \sin X \qquad y = \sin Y \qquad z = \sin Z$$
Where $X + Y + Z = \pi$ (meaning that those are the angles of a triangle).
