Prove that the sum of the medians of a triangle is greater than the semiperimeter

Question

Prove that the sum of the medians of a triangle is greater than the semiperimeter

I tried working backwards. That is I presumed that $$AF + BE + DC > \frac{AB + BC + AC}{2}$$ is true. Then I came up with $AD + BF + EC < AF + BE + DC$ at the end.

Answer 1

$M$ is the mid point of $AB$ and $m_c$ is the medians coming from $C$. We also have that $AB=c$, $BC=a$ and $AC=b$.

Then in the triangle $CMA$ we have, by triangle inequality

$$b<m_c+\frac{c}{2}\quad (1)$$

and in the triangle $CMB$ we have, by triangle inequality

$$a<m_c+\frac{c}{2}\quad (2)$$

so, $(1)+(2)$ give us,

$$a+b<2m_c+c \quad (3)$$

similarily, we have

$$a+c<2m_b+b\quad (4)\\ b+c<2m_a+a\quad (5)$$

now $(3)+(4)+(5)$ we get

$$m_a+m_b+m_c>\frac{a+b+c}{2}$$

Answer 2

You cannot prove an assertion by assuming it is true and working backwards (although you might derive a contradiction and so prove that the assertion is not true - but that is not what you are trying to achieve in this question).

Think of triangles that have a median as one side. Use the triangle inequality to derive three inequalities (one for each median) that have the form median > some expression. Then add these inequalities together to find a lower limit on the sum of the medians.

Answer 3

By the triangle inequality $$\sum_{cyc}\left(\frac{2}{3}m_a+\frac{2}{3}m_b\right)>\sum_{cyc}c.$$ Hence, $$m_a+m_b+m_c>\frac{3}{4}(a+b+c)>\frac{1}{2}(a+b+c)$$

Answer 4

Applying the triangle inequality to each blue triangle we have $$\begin{aligned}\frac23m_a+\frac13m_b&>\frac{c}{2}\\ \frac23m_b+\frac13m_c&>\frac{a}{2}\\ \frac23m_c+\frac13m_a&>\frac{b}{2} \end{aligned}$$

Now add them all up we get the desired inequality.