<s>[INST] 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.
 [/INST] With your substitution you have $a,b,c\in [0,2\pi],\ a+b+c=\pi$ and
$$\cos a+\cos b+\cos c=\frac{3}{2}$$
We have that:
$$\cos c=1-2\sin^2 \frac{c}{2},\ \cos a+\cos b=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}$$
Therefore:
$$\cos a+\cos b+\cos c=2\cos \frac{a+b}{2}\cos\frac{a-b}{2}+\cos c\\
=2\sin \frac{c}{2}\cos \frac{a-b}{2}+1-2\sin^2\frac{c}{2}\leq 2\sin \frac{c}{2}+1-2\sin^2\frac{c}{2}$$
Therefore:
$$\frac{3}{2}\leq 2\sin \frac{c}{2}+1-2\sin^2\frac{c}{2}\Rightarrow 2\left(\sin \frac{c}{2}-\frac{1}{2}\right)^2\leq 0$$
This implies $c=\frac{\pi}{3}$. Similarly we can prove that $a=b=\frac{\pi}{3}$, which gives $x=y=z=\frac{1}{2}$.
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