For any complex $z$, $|z-1|\leq |z-j|+|z-j^2|$

Question

Let $z\in \mathbb C$. Prove that $|z-1|\leq |z-j|+|z-j^2|$

This inequality appears as an exercise in a book for highschoolers. It is marked as very difficult. $j=\exp(2i\pi/3)$ denotes a third root of unity.

I tried squaring both sides, and making use of $j^2=-1-j$, as well as $\bar j=j^2$ but I haven't made significant progress...

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Remark: In retrospect it seems obvious that the problem cannot be solved without geometric insights, that I didn't know of at the time.

Answer 1

(The following is equivalent to Aqua's elegant answer. I'm just stating it to demonstrate a self-contained solution.)

Start with the identity $$ (z-1)(j-j^2) = (z-j)(1-j^2) + (z-j^2)(j-1) \, . $$ Then apply the triangle inequality and use that $$ |j-j^2| = |1-j^2| = |j-1| \, . $$

Answer 2

Let $A\sim 1$, $B\sim j$ and $C\sim j^2$ then $ABC$ is an equilateral triangle in $\mathbb{C}$. Let $D\sim z$. Then by Ptolemy's inequality we have:

$$AD\cdot BC\leq AB\cdot CD+AC\cdot BD$$

Since $BC=AB = AC$ and $AD =|z-1|$, $CD=|z-j^2|$, $BD =|z-j|$ we get what we want.