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Given $\triangle ABC$ is acute, prove that $$\cot A + \cot B + \cot C \ge \sqrt{3}$$

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    Please improve the question by adding extra context, such as what you have tried. – 2017-02-10

2 Answers 2

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We have that, since $\cot x $ is a convex function on $\left(0, \frac{\pi}{2} \right)$, and $A,B,C \in \left( 0, \frac{\pi}{2} \right)$ that $$\cot A + \cot B + \cot C \ge 3 \cot \left( \frac{A+B+C}{3} \right) =\sqrt{3}$$ Following from Jensen's Inequality. We are done!

Inequalities involving the trigonometric values of angles on a triangle normally always involve Jensen's Inequality, which is something you may want to know.

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    Hm, that inequality looks useful... – 2017-02-10
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    @SimplyBeautifulArt Useful where? – 2017-02-11
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    In inequalities of course! (and I pinged you somewhere btw) – 2017-02-11
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    @SimplyBeautifulArt Of course it is useful. It can be used to prove AM-GM very simply. It's a nice inequality that is widely applicable. And I didn't have time to see where you pinged me. – 2017-02-11
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    I pinged you in the most obvious of places for you to check. – 2017-02-11
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using Identity $$\tan A+\tan B+\tan C = \tan A \tan B\tan C$$

So $$\cot A\cot B+\cot B\cot C+\cot C \cot A = 1$$

and $$(\cot A-\cot B)^2+(\cot B-\cot C)^2+(\cot C-\cot A)^2 \geq 0$$

so $$\cot^2 A+\cot^2 B+\cot^2 C\geq 1$$

So $$(\cot A+\cot B+\cot C)^2 = \cot^2 A+\cot^2 B+\cot^2 C+2(\cot A\cot B+\cot B\cot C+\cot C \cot A)\geq 3$$

so $$\cot A+\cot B+\cot C\geq \sqrt{3}$$ and equality hold when $\cot A=\cot B=\cot C$

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    Are you sure about the identity that is used in your answer? – 2017-02-10
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    which is true when $A+B+C = \pi$ – 2017-02-10