Prove that $ \sum\limits_{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2$

Question

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.

Answer 1

Let $\cot\alpha=a$, $\cot\beta=b$ and $\cot\gamma=c$.

Hence, $ab+ac+bc=1$ and we need to prove that $\sum\limits_{cyc}\sqrt{a+b}\geq2\sqrt2$ or

$a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}\geq4$, which is true because by C-S $a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}=a+b+c+\frac{1}{2}\sum\limits_{cyc}\sqrt{(1+3)(a^2+1)}\geq$ $\geq a+b+c+\frac{1}{2}\sum\limits_{cyc}(a+\sqrt3)=\frac{3}{2}(a+b+c)+\frac{3\sqrt{3}}{2}\geq\frac{3\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}=3\sqrt3>4$

Answer 2

hint:

$\sqrt{\cot{A}+\cot{B}}=\sqrt{\dfrac{sin(A+B)}{sinAsinB}}=\sqrt{\dfrac{sinC}{sinAsinB}}=\dfrac{sinC}{\sqrt{sinAsinBsinC}}$

$3\sqrt[3]{sinAsinBsinC} \le sinA+sinB+sinC \le \dfrac{3\sqrt{3}}{2}$