Prove that the median of the largest segment is the smallest. Prove that the median of the largest segment is the smallest.
In the book it is written that it is easy provable using triangle inequality But where should I perform the triangle inequality?Any hints?
 A: Let the centroid of the triangle be at the origin, and let ${\bf a}$, ${\bf b}$, ${\bf c}$ be the position vectors of its vertices $A$, $B$, $C$. Then $${\bf a}+{\bf b}+{\bf c}={\bf 0}\ ;\tag{1}$$ furthermore the lengths of the medians are ${3\over2}|{\bf a}|$, ${3\over2}|{\bf b}|$, ${3\over2}|{\bf c}|$.
Assume that $|AB|\geq|AC|$, which is the same as $|{\bf b}-{\bf a}|\geq|{\bf c}-{\bf a}|$. Eliminating ${\bf a}$ here using $(1)$ gives $|2{\bf b}+{\bf c}|\geq|{\bf b}+2{\bf c}|$, and by squaring we obtain
$$4|{\bf b}|^2+4{\bf b}\cdot{\bf c}+|{\bf c}|^2\geq |{\bf b}|^2+4{\bf b}\cdot{\bf c}+4|{\bf c}|^2\ .$$
This implies $|{\bf b}|\geq|{\bf c}|$, as claimed.
A: I assume that $a \lt b \lt c$.
 
Use cosine law twice to setup equations for 
(1) … $(M_C)^2$ in terms of b, c, cos A.
(2) …$(M_B)^2$ in terms of b, c, cos A.
In the process of doing (1) – (2), the terms involving cosine will be cancelled out. From the assumed fact that $c \gt b$, we can arrive at the conclusion of $M_B \gt M_C$.
Note:- Like to see a solution using triangle inequality.
