# Proving $3\;|a_1-a_3|+|b_1-b_3|\leq 3\;|a_1-a_2|+|b_1-b_2|+3\;|a_2-a_3|+|b_2 - b_3|$

I'd like to prove that $$3\;|a_1-a_3|+|b_1-b_3|\leq 3\;|a_1-a_2|+|b_1-b_2|+3\;|a_2-a_3|+|b_2 - b_3|$$

Obs.: I don't know if this is possible.

### My tentative (edited)

For the left side,

$$$$3\mid (a_1 - a_2) + (a_2 - a_3) \mid + \mid(r_1 - r_2) + (r_2 - r_3) \mid,$$$$

however, I don't know how to continue...

• I think you are almost there. Just check your sign mistake. Aug 26 '20 at 3:48
• Thanks! I corrected the signals. But, I can't see it.
– WJFS
Aug 26 '20 at 3:52

Use the triangular inequality twice $$|a_2-a_3|+|a_1-a_2|\ge |a_1-a_3|$$ and a similar one for $$b$$'s.