How to show that the triangle is equilateral triangle? 
If $\cot A+\cot B+\cot C=\sqrt3$ then prove that the triangle is equilateral triangle.

Trial: I can counter check that this is true as $\cot 60+\cot 60+\cot 60=3 \frac{1}{\sqrt3}=\sqrt3$.Here I also know that $A+B+C=180$. But I am unable to show it. Please help.
 A: $$\text{Now,}\sum(\cot A-\cot B )^2=2\sum(\cot^2A-\cot A\cot B)$$
$$\text{Again,}\sum\cot^2A=(\sum\cot A)^2-2\sum \cot A\cot B$$
We have $\displaystyle\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$
and as $\displaystyle A+B+C=\pi, \cot(A+B)=\cot(\pi-C)=-\cot C$
Comparing values of $\displaystyle\cot(A+B),$ we get $\displaystyle\sum \cot A\cot B=1$
If we set $\cot A=p$ etc.,
we have $pq+qr+rp=1$ and $p+q+r=\sqrt3$
$\displaystyle\implies(p-q)^2+(q-r)^2+(r-p)^2=2(p^2+q^2+r^2-pq-qr-rp)=2\{(p+q+r)^2-2(pq+qr+rp)-pq-qr-rp\}=2\{(p+q+r)^2-3(pq+qr+rp)\}=2\{(\sqrt3)^2-3\}=0$
As $p,q,r$ are real, we have $\displaystyle p=q=r$
$\displaystyle\implies\cot A=\cot B\implies A=n\pi+B$ where $n$ is any integer.
But as $0<A,B<\pi, A=B$
A: Since $A + B + C = \pi$, $\cot C = \cot(\pi - (A + B)) = -\cot(A + B)$. Using the angle addition formula for cotangent:
$$\cot A + \cot B - \frac{\cot A \cot B - 1}{\cot A + \cot B} = \sqrt{3}$$
and if we introduce new variables such that $p = \cot A + \cot B, q = \cot A \cot B$:
$$p + \frac{1 - q}{p} = \sqrt{3}$$
$$p^2 - \sqrt{3}p + (1 - q) = 0$$
$$q = p^2 - \sqrt{3}p + 1$$
By Vieta, the roots of $x^2 - px + q = 0$ are $\cot A, \cot B$. We want these roots to be real, thus we can set a condition on the discriminant:
$$p^2 - 4q ≥ 0$$
$$p^2 - 4(p^2 - \sqrt{3}p + 1) ≥ 0$$
$$-3 \left(p^2 - \frac{4}{\sqrt 3}p + \frac{4}{3} \right) ≥ 0$$
$$-3 \left(p - \frac{2}{\sqrt 3} \right)^2 ≥ 0$$
and for this concave down quadratic, this can only occur when $p = \frac{2}{\sqrt 3}$ and $q = \frac{4}{3} - 2 + 1 = \frac{1}{3}$.
Since we already know that the quadratic is a perfect square, $\cot A, \cot B$ must be the roots of $\left(x - \frac{1}{\sqrt 3} \right)^2 = 0$, which implies $\cot A = \cot B = \frac{1}{\sqrt 3}$ or that $A = B = \frac{\pi}{3}$ as $A + B + C = \pi$.
Thus $\Delta ABC$ must be equilateral.
