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,

\begin{equation} 3\mid (a_1 - a_2) + (a_2 - a_3) \mid + \mid(r_1 - r_2) + (r_2 - r_3) \mid, \end{equation}

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

  • $\begingroup$ I think you are almost there. Just check your sign mistake. $\endgroup$
    – cr001
    Aug 26 '20 at 3:48
  • $\begingroup$ Thanks! I corrected the signals. But, I can't see it. $\endgroup$
    – 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.


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