Question

Prove that the perimeter of a triangle is greater than the sum of its three medians.

Answer 1

Let ABC be the triangle and D, E and F be the midpoints of BC, CA and AB, respectively.
Since the sum of two sides of a triangle is greater than twice the median bisecting the third side, we have:
$$\begin{gathered} AB+AC>2AD \\ \mathrm{Similarly},BC+AC>2CF \\ \mathrm{Also},BC+AB>2BE \end{gathered}$$
On adding all opf the above inequalities, we get:
$$\begin{gathered} (AB+AC)+(BC+AC)+(BC+AB)>2AD+2CD+2BE \\ \Rightarrow 2(AB+BC+AC)>2(AD+BE+CF) \\ \therefore AB+BC+AC>AD+BE+CF \end{gathered}$$

Thus, the perimeter of triangle is greater than the sum of the medians.