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 :)
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 :)
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 $x\in\left(0,\frac{\pi}{2}\right)$.