Proof of Hlawka's inequality for complex numbers.

Question

Prove that $$|z_1 + z_2| + |z_2 + z_3| + |z_3 + z_1| \le |z_1 + z_2 + z_3| + |z_1| + |z_2| + |z_3|$$

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$$\begin{align} &\ \ |z_1 + z_2 + z_3| + |z_1| + |z_2| + |z_3| \\ =&\ \ |(z_1 + z_3) + (z_2 + z_3) - z_3| + |z_1| + |z_2| + |z_3| \\\ge &\ \ |(z_1 + z_3) + (z_2 + z_3)| - |z_3| + |z_1| + |z_2| + |z_3| \\=&\ \ |(z_1 + z_3) + (z_2 + z_3)| + |z_1| + |z_2| \\\ge& \ \ |(z_1 + z_3) + (z_2 + z_3)| + |z_1 + z_2| \end{align}$$

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• I am stuck here. Please help me complete the proof.

Answer 1

We need to prove that $$\left(|a+b|+|a+c|+|b+c|\right)^2\leq\left(|a+b+c|+|a|+|b|+|c|\right)^2$$ and since $$|a+b|^2+|a+c|^2+|b+c|^2=|a+b+c|^2+|a|^2+|b|^2+|c|^2,$$ it's enough to prove that $$\sum_{cyc}\left(|a(a+b+c)|+|bc|\right)\geq\sum_{cyc}|(a+b)((a+c)|,$$ which is true because $$|a(a+b+c)|+|bc|\geq|a(a+b+c)+bc|=|(a+b)(a+c)|$$ Done!