Prove that the sum of any two medians of a triangle is greater than the third median.?

Question

I must emphasize that there is no restriction on what type of triangle it is and therefore the lengths of the medians cannot be equal.

Answer 1

Let the median of the given triangle are m1,m2 and m3.
Then m1+m2 >m3 or m2+m3>m1 or m3+m1>m1 (assume m1=5, m2=6 and m3=6)

then 5+6>6 or 6+6>5 or 6+5>6.
o.k

Answer 2

Proof. Construct a parallelogram ABCE such that triangle ABC is congruent to triangle ECB. By triangle inequality, AC + EC > 2AD (diagonals of parallelogram bisect each other) So, AB + AC > AC (AB is congruent to EC)

Answer 3

consider the length of the medians as "a"
so a+a=2a
third median =a
since 2a is greater than a
sum of two medians of a triangle is greater than the thiird

Answer 4

you have the formulas for the median's so derive them or look at the limits at infinity -infinity.