I know that $|z|\sqrt{2} \geq |Re(z)| + |Im(z)|$.
When I tried to prove it I did this:
$|z| = |Re(z) + iIm(z)|$
By the triangle inequality:
$|Re(z) + iIm(z)| \leq |Re(z)| + |iIm(z)|$
$ |Re(z)| + |iIm(z)| = |Re(z)| + |i||Im(z)| = |Re(z)| + |Im(z)|$
So far:
$|z| \leq |Re(z)| + |Im(z)|$
Multiplying by $\sqrt{2}$ on both sides:
$|z|\sqrt{2} \leq (|Re(z)| + |Im(z)|)\sqrt{2}$
Since $|Re(z)| + |Im(z)| \leq \sqrt{2}(|Re(z)| + |Im(z)|)$:
$|z|\sqrt{2} \leq |Re(z)| + |Im(z)|$
What's wrong with this?
We finally obtain
$$|z| \leq |Re(z)| + |Im(z)| \iff \sqrt{x^2+y^2}\le |x|+|y|\iff x^2+y^2\le x^2+y^2+2|xy|\iff |xy|\ge 0$$
which is true and therefore it is also true that
$$\sqrt 2|z| \leq \sqrt 2(|Re(z)| + |Im(z)|)$$
as noticed the wrong step is in the claim
$$|z| \leq |Re(z)| + |Im(z)| \quad \land \quad |Re(z)| + |Im(z)| \leq \sqrt{2}(|Re(z)| + |Im(z)|)\\\implies \sqrt 2|z| \leq |Re(z)| + |Im(z)|$$
which is not true in general, indeed note that
$$|z|\sqrt{2} \leq |Re(z)| + |Im(z)| \iff \sqrt 2 \sqrt{x^2+y^2}\le |x|+|y|\iff \sqrt{\frac{x^2+y^2}2}\le \frac{|x|+|y|}2$$
which is clearly false in general for RMS-AM inequality and only equality holds if and only if $x=y$.