Show the following equality Basically I want to show the following:
$$\sqrt{2}\ |z|\geq\ |\operatorname{Re}z| + |\operatorname{Im}z|$$
So what I did is the following:
Let $z = a + bi$
Consider the following:
$$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$$
Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$
Thus $2|z|^2 \geq a^2+b^2+2ab = (a+b)^2$
Hence $\sqrt{2}|z| \geq a + b$ but $a = |\operatorname{Re}z|~,~ b = |\operatorname{Im}z|$
Did I make a mistake somewhere, if yes I would appreciate it if it could be pointed out and perhaps provide some guideline on how to prove this.
Let $z = r\cos\theta + ri\sin\theta$

Then $|Re(z)| + |Im (z)| = \pm\sqrt 2 r\sin(\theta \pm \frac{\pi}{4}) \leq \sqrt 2 r$

All four possible sign combinations will cover $\theta$ in all quadrants.
Observe
\begin{align}
|a|\cdot 1+|b|\cdot 1 \leq \sqrt{a^2+b^2}\sqrt{1^2+1^2}.
\end{align}