Here is the proof with Motivation:\[\]Since we are asked to prove \[2/3AD+2/3B3+2/3CF<AB+BC+CA\] and \(AD,BE,CF\) are medians,it immediately comes to mind that \(2/3AD=AG\)(G:Centroid),now we need a relation between \(AG\) and \(AB\),the best way to go is obviously the triangle inequality.Hence we apply it to the triangle containing \(AG,AB\) and that is triangle \(AGB\),using the same reasoning,apply it to the triangles \(BGC,AGC\).Add the three inequalities.The second part is left as an exercise for you.Take inspiration form the first part's solution and try to do it.Oh,and BTW your question has opposite signs.Please correct it.

## Comments

Sort by:

TopNewestHere is the proof with Motivation:\[\]Since we are asked to prove \[2/3AD+2/3B3+2/3CF<AB+BC+CA\] and \(AD,BE,CF\) are medians,it immediately comes to mind that \(2/3AD=AG\)(G:Centroid),now we need a relation between \(AG\) and \(AB\),the best way to go is obviously the triangle inequality.Hence we apply it to the triangle containing \(AG,AB\) and that is triangle \(AGB\),using the same reasoning,apply it to the triangles \(BGC,AGC\).Add the three inequalities.The second part is left as an exercise for you.Take inspiration form the first part's solution and try to do it.Oh,and BTW your question has opposite signs.Please correct it.

Log in to reply

Thanks and i got the 2nd part... :)

Log in to reply

Let the medians intersect at G.

AG+BG>AB

=> 2/3

AD+2/3BE>AB.Similarly u can find the inequalities for other sides.

Now add up the inequalities.

The first half of the problem is pretty obvious.

AB+BD>AD

CD+AC>AD

Similarly find the inequalities for other sides.

I hope this was clear to u!

Log in to reply