# An Inequality with Complex Numbers of Unit Length

### What Might That Be About?

### Problem

### Solution 1

The situation is illustrated with the above applet. A worthy observation is that the condition $|a+b+c|\lt 1\,$ is equivalent to claiming that $a,b,c\,$ do not lie in a half-plane whose boundary (a straight line passes through the origin. WLOG, assume that $a=1,\,$ $b=\cos s+i\sin s,\,$ and $c=\cos t-i\sin t.\,$ The aforementioned condition of not belonging to a half-plane is expressed in terms of the following inequalities: $0\lt s,t\lt \pi\,$ and $s+t\gt\pi.$

Now, $\displaystyle |a-b|=2\sin\frac{s}{2},\,$ $\displaystyle |a+b|=2\cos\frac{s}{2},\,$ $\displaystyle |a-c|=2\sin\frac{t}{2},\,$ $\displaystyle |a+c|=2\cos\frac{t}{2}.\,$ We have to prove that

$\displaystyle \sin\frac{s}{2}+\sin\frac{t}{2}\gt \cos\frac{s}{2}+\cos\frac{t}{2}.$

Note that , for $s+t=\pi\,$ the inequality becomes equality.

We proceed with two fundamental identities:

$\displaystyle\begin{align} \sin\alpha+\sin\beta&=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \cos\alpha+\cos\beta&=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}. \end{align}$

Note that, with $0\lt s,t\lt\pi,\,$ we may be certain that $\displaystyle \frac{s-t}{4}\ne\frac{\pi}{2},$ so that $\displaystyle\cos\frac{s-t}{4}\ne 0.\,$ The inequality to prove becomes $\displaystyle\sin\frac{s+t}{4}\gt\cos\frac{s+t}{4},\,$ or $\displaystyle\tan\frac{s+t}{4}\gt 1,\,$ which is true because $\tan (x),$ is increasing on $\displaystyle \left[0,\frac{\pi}{2}\right),\,$ so that $\displaystyle\tan\frac{s+t}{4}\gt \tan\frac{\pi}{4}=1.$

### Solution 2

Let zero be represented by $O$. Let the angle that $\overrightarrow{OA}$ needs to be rotated by to align with $\overrightarrow{OB}$ be $\alpha$. The rotation could be in clockwise or anti-clockwise sense. We choose the sense that ensures $0 \leq \alpha \leq \pi$. We define $\beta$ analogously as the angle between $\overrightarrow{OA}$ and $\overrightarrow{OC}$ (with $0 \leq \beta \leq \pi$).

Thus,

$\begin{align}P &\overset{\Delta}{=} |a-b|+|a-c|-|a+b|-|a+c| \\ \nonumber &=2 \left[ \sin \left( \frac{\alpha}{2} \right) + \sin\left(\frac{\beta}{2}\right) - \cos\left(\frac{\alpha}{2}\right) -\cos\left(\frac{\beta}{2}\right) \right] \\ \nonumber &=2\sqrt{2}\left[\sin\left(\frac{\alpha}{2}-\frac{\pi}{4}\right) + \sin\left(\frac{\beta}{2}-\frac{\pi}{4}\right) \right] \\ \nonumber &=4\sqrt{2} \sin\left(\frac{\alpha+\beta}{4}-\frac{\pi}{4}\right) \cos\left(\frac{\alpha-\beta}{4}\right) \end{align}$

We need to prove that $P \geq 0$. The argument of the cosine term is limited to $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$. Thus, the cosine term is non-negative. The argument of the sine term is also limited to $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$. Thus, the sine term is negative only if $(\alpha+\beta) < \pi$ (condition Q).

The constraint can be written as

$\begin{align}&|a+b+c|^2 \leq 1 \\ \nonumber & (1+\cos\alpha+\cos\beta)^2+(\sin\alpha \pm\sin\beta)^2 \leq 1 \\ \nonumber & 3 + 2\cos\alpha + 2\cos\beta + 2\cos (\alpha \mp\beta)] \leq 1 \\ \nonumber & 3 + 4\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)+4\cos^2 \left(\frac{\alpha \mp \beta}{2}\right) -2 \leq 1 \\ \nonumber & \cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)+\cos^2 \left(\frac{\alpha \mp \beta}{2}\right) \leq 0 \\ \nonumber \end{align}$

The two possibilities $\pm$ arise because the rotational sense used in the definition of $\alpha$ and $\beta$ could be the same or opposite.

**Case 1 (Same rotational sense):**

$\begin{align}&\cos\left(\frac{\alpha-\beta}{2}\right) \left[\cos\left(\frac{\alpha+\beta}{2}\right) + \cos\left(\frac{\alpha-\beta}{2}\right) \right] \leq 0 \\ \nonumber &\cos\left(\frac{\alpha-\beta}{2}\right) \cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) \leq 0 \end{align}$

The argument of all the cosine terms lie either in the first or the fourth quadrant where the cosine is non negative. Hence, this condition is satisfied only if either or both $\alpha$ and $\beta$ equal $\pi$. Thus, condition Q is not true and $P \geq 0$ in this case.

**Case 2 (Opposite rotational sense):**

$\begin{align}&\cos\left(\frac{\alpha+\beta}{2}\right) \left[\cos\left(\frac{\alpha-\beta}{2}\right) + \cos\left(\frac{\alpha+\beta}{2}\right) \right] \leq 0 \\ \nonumber &\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) \leq 0 \end{align}$

The arguments on the second and third cosine terms lie in the first or the fourth quadrant and the condition for either of the terms being equal to zero reduces to case 1. However the first cosine term can be non-positive, even with neither $\alpha$ nor $\beta$ being equal to $\pi$, if the argument is in the second quadrant. This condition is $2\pi \gt \alpha+\beta \geq \pi$. This condition ensures that condition Q fails and $P \geq 0$ in this case.

### Solution 3

Consider \( a, b, c \) as elements of \( \mathbb{R}^2 \). Then \( \| a+b+c \| \leq 1 \) means that \( (a+b+c)\cdot a \leq 1 \), and with \( a\cdot a = 1 \) obtain \( (b+c)\cdot a \leq 0 \). Furthermore \( (b+c)\cdot b \geq 0 \) and \( (b+c)\cdot c \geq 0 \), so \( b,c,-a \) lie in the same half-space \( H_{b+c} = \{ x \mid x\cdot (b+c) \geq 0 \} \).

On the other hand, there exists a (filled) ellipse \( E = \{ x \mid \| x - b \| + \| x - c \| \leq R \} \) with focal points \( b, c \) which contains only the part of the unit circle on the half space \( H_{b+c} \). That is, \( S^1 \cap H_{b+c} = S^1 \cap E \).

If \( (b+c)\cdot a = 0 \), then \( a, -a \in \partial E \) and we have equality \( \| a - b \| + \| a - c \| = \| a + b\| + \| a + c\| \). Otherwise \( (b+c)\cdot a < 0 \) and then \( -a \in S^1 \cap H_{b+c} = S^1 \cap E \) and \( a \notin E \). In particular, \( \| a - b \| + \| a - c \| > R \geq \| - a - b \| + \| -a - c \| \).

### Illustration

### Acknowledgment

The problem which is the invention of Leo Giugiuc and Kadir Altintas, has been posted by Leo Giugiuc on the CutTheKnotMath facebook page. That's an elegantly refreshing problem. Solution 1 is by the two authors; Solution 2 s by Amit Itagi; Solution 3 is by Sophie L. Muller. The illustrated solution is by N. N. Taleb.

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