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For any acute-angled triangle $ABC$ show that

$\tan{A}+\tan{B}+\tan{C} \geq \frac{s}{r},$

where where $s$ and $r$ denote the semi-perimeter and the inradius, respectively.

Merci :)

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    $\triangle$ conventionally represents area. Please refer to http://mathworld.wolfram.com/Inradius.html.2012-09-14

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If $a$, $b$, and $c$ denote the sides of the corresponding triangle we have $s=s-a+s-b+s-c$. We also have $\frac{s-a}{r}=\cot\frac{A}{2}$ and similar identities for $b$ and $c$. So what we want to show is that $ \tan A+\tan B+\tan C\geq \cot \frac{A}{2} +\cot \frac{B}{2} +\cot \frac{C}{2}. $ This is equivalent to $\frac{1}{\cos A} + \frac{1}{\cos B}+ \frac{1}{\cos C}\geq 6.$ This inequality follows by application of the Jensen's inequality for the convex function $\frac{1}{\cos x}$ defined for $x\in\left(0,\frac{\pi}{2}\right)$.

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    If $E$ denotes the point where the incircle touches $AC$, then $AE=s-a$. Obviously, $\angle IAE = \frac{A}{2}$, $I$ being the incenter. So $\frac{s-a}{r}=\frac{AE}{IE}=\cot\frac{A}{2}$. For transition to $\sum_{cyc} \frac{1}{\cos A}$ you have to use a couple of trigonometric identities and some straightforward algebraic manipulation.2012-09-13