> 
> Let $ \triangle ABC$ be an acute-angled triangle. Prove that
>  $ \sum\limits\_\text{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2 $
> 
> 
> 

**Attempt**

Since $\triangle{ABC}$ is acute, we may say that $A,B,C \in (0, \frac{\pi}{2})$. Now, we have that by a result for triangles $\displaystyle \sum\_{cyc} \cot{A} \cdot \cot{B} = 1$. Then see that $$ \sum\_{cyc} \sqrt{\cot{A}+\cot{B}} = \sqrt{\cot{A}+\cot{B}}+\sqrt{\cot{B}+\cot{C}}+\sqrt{\cot{A}+\cot{C}}.$$ Now see that by Cauchy-Schwarz $\sqrt{(\cot{A}+\cot{B})(2)} \geq (\sqrt{\cot{A}}+\sqrt{\cot{B}})$ and thus $\sqrt{\cot{A}+\cot{B}}+\sqrt{\cot{B}+\cot{C}}+\sqrt{\cot{A}+\cot{C}} \geq \sqrt{2}(\sqrt{\cot{A}}+\sqrt{\cot{B}}+\sqrt{\cot{C}})$. Now since $$\cot{A}\cot{B}+\cot{B}\cot{C}+\cot{A}\cot{C} = 1.$$

I get stuck here.