Proving that $\sqrt{2} |z| \geq |x| + |y|$ for $z = x + iy$

Question

There is a question in my textbook:

Verify that $\sqrt{2} |z| \geq |x| + |y|$ for $z = x + iy$ with $x, y \in \mathbb{R}$.

Hint: Boil this down to $(|x|-|y|)^2 \geq 0$.

I followed the hint and logically arrived at the fact that if $\sqrt{2} |z| \geq |x| + |y|$ then $(|x|-|y|)^2 \geq 0$ is true. However, I couldn't see how it helps me prove the converse to be true:

My work:

Let $z = x+iy$, then \begin{align*} &\, \sqrt{2}|z| = \sqrt{2}\sqrt{x^2 + y^2} \geq |x| + |y| \\ \iff&\, 2x^2 + 2y^2 \geq |x|^2 + |y|^2 +2|x||y| \\ \iff&\, 2x^2 + 2y^2 - 2|x||y| \geq |x|^2 + |y|^2 \\ \iff&\, (|x| - |y|)^2 \geq |x|^2 + |y|^2 \\ \implies&\, (|x| - |y|)^2 \geq 0. \end{align*}

It seems like the last line needs an "iff" in order to be powerful in answering this problem, but I only see it going one way. Can someone please explain why it is an "iff" statement instead of just "if"?

Answer 1

$$\sqrt{2}|z| = \sqrt{2}\sqrt{x^2 + y^2} \geq |x| + |y| \\ \iff 2x^2 + 2y^2 \geq |x|^2 + |y|^2 +2|x||y| \\ \iff 2x^2 + 2y^2 - 2|x||y| \geq |x|^2 + |y|^2 $$

At this point cancel $x^2+y^2$ from both sides to get

$$\iff x^2 + y^2 - 2|x||y| \geq 0 $$

$$\iff (|x| - |y|)^2 \geq 0 $$

And you are done.

Answer 2

We have that

$$\sqrt{2}\sqrt{x^2 + y^2} \geq |x| + |y| \iff \sqrt{\frac{x^2 + y^2}2} \geq \frac{|x| + |y|}2$$

which is true by RMS-AM inequality.