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.

inequality between the mdeians and exradii

Proof

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, $m_a\ge l_a:$

Median longer than angle bisector

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.

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