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 $z=a+bi$. Then $|z|^2=a^2+b^2$ and Re$z=a$ and Im$z=b$. Observe that
$$(\sqrt{2}|z|)^2=2(a^2+b^2)$$ and $$\Big(|\text{Re}z|+|\text{Im}z|\Big)^2=a^2+2|ab|+b^2.$$
Now,
$$
\begin{align}
(|a|-|b|)^2\geq 0&\iff a^2+b^2-2|ab|\geq 0\\
&\iff2(a^2+b^2)\geq a^2+2|ab|+b^2\\
&\iff (\sqrt{2}|z|)^2\geq\Big(|\text{Re}z|+|\text{Im}z|\Big)^2\end{align}
$$
Since $\sqrt{2}|z|$ and $|\text{Re}z|+|\text{Im}z|$ are nonnegative, the result follows.
