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.

Another way to solve the same could be $\cos^{-1}x=A$, $\cos^{-1}y=B$ and $\cos^{-1}z=C$ 
and thus $A+B+C=\pi$ and the condition becomes $\cos A+\cos B+\cos C=\frac{3}{2}$, 
which can be simplified to $2\cos\frac{A+B}{2}\frac{A-B}{2}+1-2\sin^2{\frac{c}{2}}=\frac{3}{2}$, and we know $\cos\frac{A+B}{2}=\cos\frac{\pi-C}{2}=\sin\frac{C}{2}$
Rearranging as quadratic in $\sin\frac{c}{2}$ we can write the equation as $2\sin^2\frac{c}{2}-2\cos\frac{A-B}{2}\sin\frac{c}{2}+\frac{1}{2}=0$
For real roots $D\ge0$, so we get $4\cos^2\frac{A-B}{2}-4\ge0$
i.e. $\sin^2\frac{A-B}{2}\le0$ which is only possible when $\sin^2\frac{A-B}{2}=0$
Therefore, $A=B=C=\frac{\pi}{3}\Rightarrow x=y=z=\frac{1}{2}$ 
