An Inequality in Triangle, II

Let, as usual, $m_a,\;m_b,\;m_c,\;l_a,\;l_b,\;l_c$ denote the lengths of the medians and the angle bisectors in $\Delta ABC,$ with sides $a,\;b,\;c$ and semiperimeter $p.$ Then

$m_a l_a+m_b l_b+m_c l_c\ge p^{2}.$

Proof

As an auxiliary result, observe that, for two vectors $\overrightarrow{u}$ and $\overrightarrow{v},$

$\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\ge \left|\overrightarrow{u}\cdot\overrightarrow{v}\right|\ge \overrightarrow{u}\cdot\overrightarrow{v},$

where the dot denotes the scalar product.

Now, let $M$ be the midpoint of $BC,$ $D$ the foot of the angle bisector at $A.$ We have

$\begin{align}\displaystyle m_a l_a &= \left|\overrightarrow{AM}\right|\left|\overrightarrow{AD}\right|\\ &=\left|\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}\right|\left|\frac{b\overrightarrow{AB}+c\overrightarrow{AC}}{b+c}\right|\\ &\ge\left(\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}\right)\cdot\left(\frac{b\overrightarrow{AB}+c\overrightarrow{AC}}{b+c}\right)\\ &=\frac{bc(b+c)+(b+c)\overrightarrow{AB}\cdot\overrightarrow{AC}}{2(b+c)}\\ &=\frac{(b+c)^{2}-a^{2}}{4}\\ &=p(p-a). \end{align}$

Similarly, $m_b l_b\ge p(p-b)$ and $m_c l_c\ge p(p-c).$ Adding up the three inequalities we obtain

$m_a l_a+m_b l_b+m_c l_c\ge p(p-a)+p(p-b)+p(p-c)=p^2.$

Acknowledgment

I am in debt to Leo Giugiuc for sending me that inequality with a proof. Leo credits the result to Laurentiu Panaitopol who posed the problem in 1981 in the "Gazeta Matematica".

|Contact| |Front page| |Contents| |Inequalities| |Geometry| |Up| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62050435

Search by google: