Why isn't it true that $|z|\sqrt{2} \leq |Re(z)| + |Im(z)|$?

Question

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?

Read the last few lines again. You are basically trying to say that $,a \le c$ and $b \le c \implies a \le b,$. That's obviously false. — dxiv, Aug 31 '18 at 23:56
Proof of true statement.: Let $z=a+bi$. Theorem $\sqrt{2(a^2+b^2)}\ge |a|+|b|$ Square both sides. to show $2(a^2+b^2)\ge a^2+2|ab|+b^2$ or $a^2+b^2\ge 2|ab|$ or $(|a|-|b|)^2\ge 0$.which is true. — herb steinberg, Sep 01 '18 at 00:09

user · Accepted Answer · 2018-09-01T01:24:41.550

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 $x=y$.

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