Show that $x=y=z$ where $\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi$ Given that $$\cos^{-1} x+ \cos^{-1} y + \cos^{-1} z = \pi.$$ Also given that $$x+y+z=\frac{3}{2}.$$ Then prove that $x=y=z.$
My attempt: Let us assume $$\cos^{-1} x=a,\> \cos^{-1} y =b, \> \cos^{-1} z=c.$$ Then we have $$a+b+c=\pi \implies a+b = \pi - c.$$ This follows that \begin{align*} \cos(a+b)=\cos(\pi - c) & \implies \cos a \cos b - \sin a \sin b = - \cos c \\ & \implies xy-\sqrt{1-x^2} \sqrt{1-y^2}=z.\end{align*} Now i am not able to proceed from here. Please help me to solve this.

This is equivalent to the equation 
of the sum of cosines of the tree angles of triangle,
\begin{align} 
\cos \alpha+\cos\beta+\cos\gamma
&=\frac{3}{2}
,
\end{align} 
it is well-known that this sum can be expressed
in terms of $r$ and $R$, 
the radii or inscribed and circumscribed circle, respectively as
\begin{align} 
\cos\alpha+\cos\beta+\cos\gamma
&=\frac rR+1=\frac{3}{2}
,
\end{align} 
which gives 
\begin{align}
R&=2\,r
,
\end{align}
and this can be true only for 
the equilateral triangle.
