Vtyjkyuk

There is a question in my textbook:

Verify that $sqrt2 |z| geq |x| + |y|$ for $z = x + iy$ with $x, y in mathbbR$.

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

I followed the hint and logically arrived at the fact that if $sqrt2 |z| geq |x| + |y|$ 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
beginalign*
&, sqrt2|z| = sqrt2sqrtx^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.
endalign*

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"?

$$sqrt2|z| = sqrt2sqrtx^2 + y^2 geq |x| + |y| \
iff 2x^2 + 2y^2 geq |x|^2 + |y|^2 +2|x||y| \
iff 2x^2 + 2y^2 - 2|x||y| geq |x|^2 + |y|^2 $$

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.

We have that

$$sqrt2sqrtx^2 + y^2 geq |x| + |y| iff sqrtfracx^2 + y^22 geq fracx2$$

which is true by RMS-AM inequality.