In a $\triangle ABC,$ If $\cot A+\cot B+\cot C =\sqrt{3},$ Then prove that $\triangle$ is equilateral. In a $\triangle ABC,$ If $\cot A+\cot B+\cot C =\sqrt{3},$ Then prove that $\triangle$ is equilateral.


$\bf{My\; Try::}$ Using Jensen's Inequality,

Let $f(x)=\cot x\;,$ Where $x\in (0,\pi),$ Then $f'(x) = -\csc^2 x$ and $f''(x) = 2\csc^2 x\cdot \cot x$

So we get $\displaystyle f''(x) = \frac{2\cos x}{\sin^3 x}&gt;0$ in $\displaystyle x\in (0,\pi)-\left\{\frac{\pi}{2}\right\}$

So $$\displaystyle\frac{\cot A+\cot B+\cot C}{3}\geq \cot\left(\frac{A+B+C}{3}\right)=\frac{1}{\sqrt{3}}$$

So we get $\cot A+\cot B+\cot C\geq \sqrt{3}$

But i did not understand how can we prove that $\triangle$ are equilateral

Thanks
You have completed the sum almost all by yourself. That's a really good try using Jensen's inequality.

For a real convex function φ, numbers $x_1$, $x_2$, ...,$ x_n$ in its domain, and positive weights $ai$, Jensen's inequality can be stated as:
$$\varphi\left(\frac{\sum a_i x_i}{\sum a_i}\right) \le \frac{\sum a_i \varphi (x_i)}{\sum a_i} \qquad\qquad(1)$$ 
and the inequality is reversed if φ is concave, which is
$$\varphi\left(\frac{\sum a_i x_i}{\sum a_i}\right) \geq \frac{\sum a_i \varphi (x_i)}{\sum a_i}.\qquad\qquad(2) $$
Equality holds if and only if $x_1=x_2=\cdots =x_n$ or $φ$ is linear.

It is given here that $$\cot A+\cot B+\cot C=\sqrt3$$

Now, observe that $\cot x$ is not linear.
So the equality in Jensen's inequality holds only if 
$$\cot A=\cot B=\cot C$$ or in other words, $$A=B=C$$
It is trivial to conclude now that $A=B=C=60^\circ$ and hence the triangle is equilateral.
You have completed the sum almost all by yourself. That's a really good try using Jensen's inequality.

For a real convex function φ, numbers $x_1$, $x_2$, ...,$ x_n$ in its domain, and positive weights $ai$, Jensen's inequality can be stated as:
$$\varphi\left(\frac{\sum a_i x_i}{\sum a_i}\right) \le \frac{\sum a_i \varphi (x_i)}{\sum a_i} \qquad\qquad(1)$$ 
and the inequality is reversed if φ is concave, which is
$$\varphi\left(\frac{\sum a_i x_i}{\sum a_i}\right) \geq \frac{\sum a_i \varphi (x_i)}{\sum a_i}.\qquad\qquad(2) $$
Equality holds if and only if $x_1=x_2=\cdots =x_n$ or $φ$ is linear.

It is given here that $$\cot A+\cot B+\cot C=\sqrt3$$

Now, observe that $\cot x$ is not linear.
So the equality in Jensen's inequality holds only if 
$$\cot A=\cot B=\cot C$$ or in other words, $$A=B=C$$
It is trivial to conclude now that $A=B=C=60^\circ$ and hence the triangle is equilateral.