How to prove that this triangle is equilateral?

Question

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.

Answer 1

Or $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=0.$$ Now, let $a\geq b\geq c$.

Thus, $$0=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-a^2b-ab^2+abc)=\sum_{cyc}a(a-b)(a-c)\geq$$ $$\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0,$$ which says that the equality occurs for $a=b=c$.

Answer 2

I Assume that all angles are acute. Now its wise to use concavity of $\cos$ to say (from Jensen's Inequality)

$$\cos(A)+\cos(B) + \cos(C) \ge 3\cos(\frac{A+B+C}{3}) = 3 \cos(\frac{\pi}{3})=\frac{3}{2}$$

equality occurs for $A=B=C = \pi / 3$.

I don't know how to handle obtuse triangles with this..,

Answer 3

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.