Question

If ${\left({\sin }^{-1}x\right)}^2+{\left({\sin }^{-1}y\right)}^2+{\left({\sin }^{-1}z\right)}^2=\frac{3{\pi }^2}{4}$, then find the minimum value of $x+y+z$

Answer 1

${\left({\sin }^{-1}x\right)}^2+{\left({\sin }^{-1}y\right)}^2+{\left({\sin }^{-1}z\right)}^2=\frac{3{\pi }^2}{4}$ Is possible only when ${\left({\sin }^{-1}x\right)}^2={\left({\sin }^{-1}y\right)}^2={\left({\sin }^{-1}z\right)}^2=\frac{{\pi }^2}{4}$
$\Rightarrow x=y=z=\pm 1\therefore min(x+y+z)=-1-1-1=-3$