AD, BE and CF are three medians of ΔABC. Let us prove that

(i) 4 (AD + BE + CF) > 3 (AB + BC + CA)

(ii) 3(AB + BC + CA) > 2(AD + BE + CF)

(i)

As we know, if G is centroid, and centroid divides the median in 2:1

Then BE = BG and FC = CG

In

Δ BGC,

BG + GC > BC (sum of two sides greater than 3^rd side)

⇒

⇒ 2BE + 2FC > 3BC

Or

3BC < 2BE + 2FC ….(1)

Similarly,

3CA < 2CF + 2AD ….(2)

And 3AB < 2AD + 2BE ….(3)

Add (1), (2) and (3)

We get,

3 BC + 3CA + 3AB < 2BE + 2FC + 2CF + 2AD + 2AD + 2BE

⇒ 3 (AB + BC + CA) = 4 BE + 4 FC + 4 AD

Or 4 (AD + BE + CF) > 3 (AB + BC + CA)

Hence proved.

(ii) In ΔACD

AC + CD > AD (sum of two sides greater than 3^rd side)

Since, AD is median

……[1]

Similarly,

In ΔABE

……[2]

In ΔBFC

……[3]

Adding [1], [2] and [3], we get

Hence, Proved!