Prove that in a triangle : $\cot A + \cot B + \cot C \ge \sqrt{3}$

Question

$\textbf{Question :}$ Prove that in a triangle $$\cot A + \cot B + \cot C \ge \sqrt{3}$$

$\textbf{My Attempt :}$

For a acute triangle I can say that all angles are less than $\frac{\pi}{2}$ and as in $[0,\frac{\pi}{2}]$ the $\cot x$ function is concave upwards we can say through Jensen's Inequality that $$\frac{\sum \cot A}{3} \ge \cot\left(\frac{\sum A}{3}\right) \implies \sum \cot A \ge \sqrt{3}$$

And now for the proof in a obtuse angled triangle, where two angles are less than $\frac{\pi}{2}$ and one is more than $\frac{\pi}{2}$. We can take each angle as an $x$-coordinate and plot the corresponding $y$-coordinate which is nothing but $(x , \cot x)$ we get three points. Now if we form a triangle with these three points, how to show that the centroid of this triangle still stays above the graph of $y=\cot x$. In the case of acute triangle it was trivial to show that but in obtuse triangle one point will have a negative $y$-coordinate. So what to do in this case?

Answer 1

I wanted to do it through the jensen method

Using Jensen's inequality for the two smaller angles $A,B \in [0, \pi/2]\,$:

$$\frac{\cot A + \cot B}{2} \ge \cot\frac{A + B}{2} = \tan \frac{C}{2}$$

Let $t = \tan \dfrac{C}{2} \gt 0$, then $\,\cot C = \dfrac{1-t^2}{2t}\,$, and (using AM-GM for the last step):

$$ \require{cancel} \cot A + \cot B + \cot C \ge 2t + \frac{1-t^2}{2t} = \frac{3}{2} t + \frac{1}{2t} \ge \bcancel{2} \sqrt{\frac{3}{\bcancel{2}} \cancel{t} \cdot \frac{1}{\bcancel{2} \cancel{t}}} = \sqrt{3} $$

Answer 2

You can make use of the fact that $\cot{A} \cot{B} + \cot{A} \cot{C} + \cot{B} \cot{C} = 1$ for any triangle $ABC$ and $x^2 + y^2 + z^2 \ge xy + xz + yz$ for all real $x,y,z$. We have

\begin{align} (\cot{A} + \cot{B} + \cot{C})^2 &= \cot^2 {A} + \cot^2 {B} + \cot^2 {C} + 2(\cot{A} \cot{B} + \cot{A} \cot{C} + \cot{B} \cot{C}) \\ &\ge 3(\cot{A} \cot{B} + \cot{A} \cot{C} + \cot{B} \cot{C}) \\ &= 3 \end{align}