<s>[INST] Prove that $ \sum\limits_{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2$ 
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.
 [/INST] hint:
$\sqrt{\cot{A}+\cot{B}}=\sqrt{\dfrac{sin(A+B)}{sinAsinB}}=\sqrt{\dfrac{sinC}{sinAsinB}}=\dfrac{sinC}{\sqrt{sinAsinBsinC}}$
$3\sqrt[3]{sinAsinBsinC} \le sinA+sinB+sinC \le \dfrac{3\sqrt{3}}{2}$
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