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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.

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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, $A=B=C=\frac{\pi}{3}\Rightarrow x=y=z=\frac{1}{2}$

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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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  • $\begingroup$ Sir this solution is fine, but I want a basic algebraic solution without Jensem's inequality. $\endgroup$ – abcdmath Mar 8 '20 at 7:53
  • $\begingroup$ @abcdmath, I have edited the solution. $\endgroup$ – LHF Mar 8 '20 at 8:01
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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.

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