How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$\therefore\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$
I have tried simplifying the given equation using cosine rule but could not get far. Please help.

A slick way is to combine Carnot's theorem with Euler's inequality $R\geq 2r$.
$$ R\cos A+R\cos B + R\cos C $$
is the sum of the signed distances of the circumcenter from the triangle sides, and it equals $R+r$. Since $r<\frac{R}{2}$ unless $ABC$ is equilateral,
$$ R\cos A+R\cos B + R\cos C = \frac{3}{2}R$$
implies that $ABC$ is equilateral.
