If $\cos A+\cos B+\cos C=\frac{3}{2}$, prove $ABC$ is an equilateral triangle If in $\Delta ABC$ $$\cos A+\cos B+\cos C=\frac{3}{2}$$prove that it is an equilateral triangle without using inequalities. I tried  using the cosine rule as follows:
$$\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$$ $\implies$
$$a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2)=3abc$$ $\implies$
$$(a+b+c)(a^2+b^2+c^2)=2(a^3+b^3+c^3)+3abc$$ 
Any clues as to how I can take it from here?

A different kind of trick : I thought you will like it.
So multiply the above equation by  $a,b,c$ and add these up. Next , use the identity that $c = a \cos B + b \cos A$, and likewise for the other sides. After pairing these, you get:
()
$$
\frac{3(a+b+c)}{2} = a \cos A + b \cos B + c \cos C + a + b +c
$$
which then simplifies to:
$$
\frac{a+b+c}{2} = a\cos A + b\cos B + c \cos C \leq \frac{(a+b+c)(\cos A + \cos B + \cos C)}3
$$
by Chebyshev's inequality. But then, equality is attained here : this happens only under certain conditions. I leave you to conclude.
