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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.

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2 Answers 2

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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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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 $a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}=a+b+c+\frac{1}{2}\sum\limits_{cyc}\sqrt{(1+3)(a^2+1)}\geq$ $\geq a+b+c+\frac{1}{2}\sum\limits_{cyc}(a+\sqrt3)=\frac{3}{2}(a+b+c)+\frac{3\sqrt{3}}{2}\geq\frac{3\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}=3\sqrt3>4$

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  • $\begingroup$ How are $\sum\limits_{cyc}\sqrt{a+b}\geq2\sqrt2$ and $a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}\geq4$ the same? $\endgroup$
    – Puzzled417
    Jan 18, 2016 at 15:52
  • $\begingroup$ @Puzzled417 $\left(\sum\limits_{cyc}\sqrt{a+b}\right)^2\geq8\Leftrightarrow2(a+b+c)+2\sum\limits_{cyc}\sqrt{(a+b)(a+c)}\geq8\Leftrightarrow a+b+c+\sum\limits_{cyc}\sqrt{a^2+1}\geq4$ $\endgroup$
    – Michael Rozenberg
    Jan 18, 2016 at 15:59
  • $\begingroup$ How does$ a+b+c+\frac{1}{2}\sum\limits_{cyc}\sqrt{(1+3)(a^2+1)} \geq a+b+c+\frac{1}{2}\sum\limits_{cyc}(a+\sqrt3)$ follow from Cauchy-Schwarz? $\endgroup$
    – Puzzled417
    Jan 18, 2016 at 16:04
  • $\begingroup$ Because $(1+3)(a^2+1)\geq(a+\sqrt3)^2$. We know that C-S it's $(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2$. $\endgroup$
    – Michael Rozenberg
    Jan 18, 2016 at 21:21
  • $\begingroup$ Where did the part of the acute-angled triangle come into your solution? $\endgroup$
    – Puzzled417
    Jan 18, 2016 at 23:10

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