Prove that in a triangle, the length of every median is smaller than the average of the lengths of the two sides meeting at the median's vertex.

Given: Consider a triangle ABC in which AD is median drawn from vertex A.

To prove: AB + AC > AD

Proof: In Δ A B D

AB + B D> AD ⇒[In a triangle sum of lengths of two sides is greater than the third side] ... (1)

In Δ A CD

AC + DC > AD [In a triangle sum of lengths of two sides is greater than the third side] ... (2)

Adding (1) and (2)

AB + AC + B D + D C > B D + DC

A B + A C + B C > 2 B D ... (3)

Also, Considering Δ AB C

AB + B C > B C ⇒[In a triangle sum of lengths of two sides is greater than the third side]

⇒ AB + B C - B C >0 ... (4)

Adding (3) and (4)

A B + B C+B C + AB + A C - B C > 2 A D

⇒2 AB + 2 A C> 2 A D

Dividing both side of inequality by 2, we get

A B + A C> A D

Hence proved.