I got this question from the first chapter of Courant and John's Introduction to Calculus and Analysis I. The problem is as follows:

Prove that $|x-a_1|+|x-a_2|+|x-a_3|\geq a_3 - a_1$, for $a_1<a_2<a_3$. For what value of $x$ does equality hold?

If you draw the "real number line"/"number axis", and plot $a_1,a_2,a_3$ and try placing $x$ into random places, you can easily see that the inequality always holds, and that equality happens only if $x=a_2$. In fact, I've proved it algebraically, only I had to divide it into 7 cases and prove each one individually (which makes the problem much easier since it allows you to get rid of every absolute value sign, but on the other hand is an extremely ugly way to solve it).

My question: is there a way to prove this directly, without dividing it into cases?

I ask this not only because I have a feeling that there is one, but also because the next problem of the book is a generalization of this, and dividing it into cases would be unfeasible, so a more direct proof of this weaker statement would certainly help me prove the more general inequality. I've tried applying the Triangle Inequality to the LHS to try and turn it into the RHS, but I couldn't finish it.


1 Answer 1


Hint: \begin{align} a_3-a_1&=|a_3-a_1|\\&=|a_3-x-(a_1-x)|\\&\leq |x-a_3|+|x-a_1|\\&\leq |x-a_1|+|x-a_2|+|x-a_3|. \end{align}


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