Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I recently saw the following inequality for complex numbers:

If $a,b\in\mathbb C$ and $|a + b|$ and $|a-b|$ are each less than or equal to 1, then

$$|a| + |b^2|/2 \leq 1.$$

How can one prove this?

share|cite|improve this question
up vote 5 down vote accepted

First, $|b| \leq 1$, since

$$ 2|b| = |b + a + b - a| \leq |b + a| + |b - a| \leq 2, $$

as noted by DonAntonio.

The conditions $|a + b| \leq 1$ and $|a - b| \leq 1$ imply that $a$ is in the intersection of the closed balls of radius $1$ centered at $b$ and $-b$, which is the shaded region here:

enter image description here

Since we essentially only care about the maximum modulus of $a$, we may rotate this region about the origin:

enter image description here

The two circles intersect at $z = \pm \sqrt{1 - |b|^2}$, which implies $|a| \leq \sqrt{1 - |b|^2}$.

Then, since the map $x \mapsto \sqrt{1-x^2} + x^2/2$ decreases from $1$ to $1/2$ in the interval $[0,1]$, we have

$$ |a| + |b|^2/2 \leq \sqrt{1 - |b|^2} + |b|^2/2 \leq 1. $$

share|cite|improve this answer

$$2|a|=|a+b+a-b|\leq|a+b|+|a-b|\leq 2\Longrightarrow |a|\leq 1$$ and of course the same's true for $\,|b|\,$ , so now the inequality's trivial.

share|cite|improve this answer
I don't think this is complete. Just having $|a| \leq 1$ and $|b| \leq 1$ does not imply $|a| + |b|^2/2 \leq 1$. For example, let $a = 1$ and $b = 1$. – Antonio Vargas May 15 '12 at 2:16
I misread and thought it was $\,\frac{|a|+|b^2|}{2}\leq 1$ . Indeed, the above isn't enough/ – DonAntonio May 15 '12 at 3:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.