An Inequality in Triangle, V
The inequality below establish a relation between the medians and the exradii of a triangle. The inequality was posted at the mathematical inequalities facebook group by Daniel Culea who found it in the book 360 Problems for Mathematical Contests by Titu Andreescu and Dorin Andrica. I am grateful to Leo Giugiuc for reposting the problem at the CutTheKnotMath facebook page and communicating to me a brilliant solution he and Dan Sitaru came up with.
We invoke a result proved previously, viz., $m_al_a\ge p(p-a),\;$ where $l_a\;$ is the length of the bisector of angle $A,\;$ and $p\;$ the semiperimeter of $\Delta ABC.\;$ Similarly $m_bl_b\ge p(p-b)\;$ and $m_cl_c\ge p(p-c).\;$
In any triangle and at any vertex the median is farther away from the altitude than the angle bisector, showing that, say, $m_a\ge l_a:$
Thus, by Heron's formula,
$m_a^2m_b^2m_c^2\ge m_al_a\cdot m_bl_b\cdot m_cl_c\ge p^3(p-a)(p-b)(p-c)=p^2S^2,$
where $S=[\Delta ABC],\;$ the area of $\Delta ABC.$
On the other hand, $r_ar_br_c=pS,\;$ which proves the required inequality.
Copyright © 1996-2018 Alexander Bogomolny