# An inequality for two complex numbers

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?

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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:

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

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

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

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

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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