# How does one prove $\sqrt{2}|z|\geq |\operatorname{Re}z|+|\operatorname{Im}z|$ holds for all $z\in \mathbb{C}?$

This is Exercise EP $14$ from Fernandez and Bernardes's book Introdução às Funções de uma Variável Complexa (in Portuguese). The authors ask us to prove that the inequality $$\sqrt{2}|z|\geq |\operatorname{Re}z|+|\operatorname{Im}z|$$ holds for all $z\in \mathbb{C}.$

I know that $|z|\geq\max\{|\operatorname{Re}z|,|\operatorname{Im}z|\}$, but all I've managed to do is show the obvious: $2|z|\geq |\operatorname{Re}z|+|\operatorname{Im}z|.$

I would appreciate a hint here.

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Please don't use display math ($$...$$) in titles. :) –  cardinal Mar 4 '12 at 21:56
@cardinal: I won't do it anymore! Thanks for helping me to do the right thing! –  spohreis Mar 4 '12 at 22:17
It happens pretty often. It's not a big deal, so I went ahead and edited it. Cheers. :) –  cardinal Mar 4 '12 at 22:20

Let $z=x+iy$, i.e. $x=\operatorname{Re}z$ and $y=\operatorname{Im}z$. Then $$(\sqrt{2}|z|)^2=2|z|^2=2x^2+2y^2.$$ On the other hand, $$(|\operatorname{Re}z|+|\operatorname{Im}z|)^2=(|x|+|y|)^2=x^2+y^2+2|xy|.$$ This implies that $$(\sqrt{2}|z|)^2-(|\operatorname{Re}z|+|\operatorname{Im}z|)^2=x^2+y^2-2|xy|=(|x|-|y|)^2\geq 0$$ which implies that $$\sqrt{2}|z|\geq |\operatorname{Re}z|+|\operatorname{Im}z|$$ as required.

Remarks: This also shows that the equality holds if and only if $|\operatorname{Re}z|=|\operatorname{Im}z|$

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