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

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 $\displaystyle f''(x) = \frac{2\cos x}{\sin^3 x}>0$ 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

## marked as duplicate by lab bhattacharjee, S.C.B., Watson, drhab, Macavity inequality StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 29 '16 at 14:17

• Maybe this will help : If a triangle is equilateral, each of its angles are equal to 60° – Tony Barbé Mar 29 '16 at 13:24
• Hint: consider when you get an equality in Jensen's inequality. – Wojowu Mar 29 '16 at 13:24
• Thanks lab bhattacharjee, I have seen that, but i want to solve it using Jenson inequality,but i did not understand why not $f''(x)>0$ for all $x\in (0,\pi)$ , here i have mention that $f''(x)=0$ at $x=\frac{\pi}{2}$, – juantheron Mar 29 '16 at 13:27
• Suppose $x \in (\frac{\pi}2, \pi)$, then $\cos x < 0, \sin x > 0 \implies f''(x) < 0$ so it's not straightforward Jensen is it? – Macavity Mar 29 '16 at 14:16

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.

• @Macavity Yes, you are right. But the equality condition remains the same in both the cases. – SchrodingersCat Mar 29 '16 at 14:21
• My point is not about the equality condition. My point is Jensen inequality is not applicable at all if the function is convex in part of the interval and concave in another part. So while I am not downvoting, the whole proof is wrong. – Macavity Mar 29 '16 at 14:23
• @Macavity Why can't we divide the interval into 2 parts? $(0,\frac{\pi}{2}]$ and $[\frac{\pi}{2},\pi)$. And then separately apply the inequality. That should be enough I hope... :-) – SchrodingersCat Mar 29 '16 at 14:25
• Because the inequality reverses when the function is concave in part of the intervale. And here you have two points where it is convex and one in which it could be concave. There are workarounds for simple cases, but it clearly is far away from the traditional Jensen inequality. – Macavity Mar 29 '16 at 14:33
• @Macavity I understand your point. But please note what I said in my first comment, the inequality might reverse in either intervals, still I can divide the interval into 2 parts, apply the inequality in both separately and yet the $\mathbf{equality}$ $\mathbf{ condition}$ remains the same and hence the conclusion can be drawn. – SchrodingersCat Mar 29 '16 at 14:37