An Inequality with Complex Numbers of Unit Length II

What Might That Be About?


An  Inequality with Complex Numbers of Unit Length II

Solution 1

To start with, let's simplify the problem by using the fact that the given numbers have unit length. Thus,

$\displaystyle\begin{align}|b+c| &= \left|a\left(\frac{b}{a}+\frac{c}{a}\right)\right|\\ &=|a|\cdot\left|\left(\frac{b}{a}+\frac{c}{a}\right)\right|\\ &=\left|\left(\frac{b}{a}+\frac{c}{a}\right)\right|\\ \end{align}$

And, similarly, $\displaystyle |a^2+bc|=\left|1+\frac{b}{a}\cdot\frac{c}{a}\right|.\,$ Now let $\displaystyle u=\frac{b}{a}\,$ and $\displaystyle v=\frac{c}{a}.\,$ Note that $|u|=|v|=1.\,$ We thus have to prove a simplified inequality:

$|1+uv|\ge |u+v|,$

subject to $|1+u+v|\le 1.$

The original situation is illustrated with the above applet. On a sister page, it was observed that the condition $|a+b+c|\le 1,$ is equivalent to $\Delta abc\,$ not being obtuse. In the applet, $a+b+c=s.$ Dividing by $a\,$ is equivalent to a rotation that maps $a\,$ to $1.$

An  Inequality with Complex Numbers of Unit Length II, proof, part 1

Now, observe that, by the definition of the product of complex numbers, $\angle (uv)0u=\angle v01\,$ and also $\angle v0(uv)=\angle 10u.\,$ It follows that angles $u0v\,$ and $10(uv)\,$ share angle bisector. Since all vectors involved have unit length, both $u+v\,$ and $1+uv$ lie on that straight line.

An  Inequality with Complex Numbers of Unit Length II, proof, part 2

Thus what needs to be shown is that $\angle u0v=\angle (-u)0(-v)\ge \angle 10(uv).$

But, since $\Delta 1uv\,$ is not obtuse, $1\,$ lies within the sector $(-u)0(-v).\,$ Now, from the fact that the angles $(-u)0(-v)\,$ and $10(uv)\,$ share the angle bisector it follows that the same is true for $uv,\,$ i.e., that $uv\,$ lies in the sector $(-u)0(-v),\,$ which completes the proof.

Solution 2

We can use polar form of the complex numbers by Euler's identity. Knowing $|a|=|b|=|c|=1,\,$ $a=e^{it},\,$ $b=e^{i(t+u)},\,$ $c=e^{i(t+u+v)}.\,$ Given that $e^{it}+e^{i(t+u)}+e^{i(t+u+v)}|\le 1,$ using complex conjugation:

$(e^{it}+e^{i(t+u)}+e^{i(t+u+v)})(e^{-it}+e^{-i(t+u)}+e^{-i(t+u+v)})\le -1,$


$e^{iu}+e^{-iu}+e^{iv}+e^{-iv}+e^{i(u+v)}+e^{-i(u+v)}+3\le -1,$

and, finally,


$\cos (u)+\cos(v)+\cos(u+v)\le 1.$

Similarly, the inequality $|a^2+bc|\ge |b+c|,\,$ can be rewritten as


$\cos (2u+v)\ge\cos(v).$

Finally, for $0\le u\le 2\pi\,$ and $0\le v\le 2\pi,\,$ condition (1) holds when

  • $u\le\pi,\,v\le\pi,\,u+v\ge\pi\,$ or
  • $u\ge\pi,\,v\ge\pi,\,u+v\le 3\pi,$

while the condition (2) holds when

  • $u\le\pi,\,u+v\ge\pi\,$ or
  • $u\ge\pi,\,u+v\le 3\pi.$

comparing regions for two inequalities

Solution 2

Noting that the constraints and the inequalities in the problem remain invariant when $a$, $b$, and $c$ are multiplied by any constant unit magnitude complex number, we, WLOG, let


The inequality constraint:

$\displaystyle \begin{align} &(a+b+c)(a^*+b^*+c^*) \leq 1 \\ &(1+e^{i\theta}+e^{i\phi})(1+e^{-i\theta}+e^{-i\phi}) \leq 1 \\ &3+2\cos\theta+2\cos\phi+2\cos(\theta-\phi) \leq 1 \\ &(\cos\theta+\cos\phi)+[1+\cos(\theta-\phi)] \leq 0 \\ &(\cos\theta+\cos\phi)+[1+\cos\theta\cos\phi+\sin\theta\sin\phi] \leq 0 \\ &(1+\cos\theta)(1+\cos\phi)+\sin\theta\sin\phi \leq 0 \\ & 4\cos^2\left(\frac{\theta}{2}\right)\cos^2\left(\frac{\phi}{2}\right)+ \sin\theta\sin\phi \leq 0 \\ &\Rightarrow \sin\theta\sin\phi\leq 0 \end{align}$

The inequality to be proven:

$\displaystyle \begin{align} &(a^2+bc)(a^{*2}+b^*c^*) \geq (b+c)(b^*+c^*) \\ &(1+e^{i(\theta+\phi)}) (1+e^{-i(\theta+\phi)}) \geq (e^{i\theta}+e^{i\phi}) (e^{-i\theta}+e^{-i\phi}) \\ & 2+2\cos(\theta+\phi) \geq 2+2\cos(\theta-\phi) \\ & -\sin\theta \sin\phi \geq \sin\theta \sin\phi \\ & \sin\theta\sin\phi \leq 0 \end{align}$


Could be looked at as a solution by cheating via computational mathematics or an exploration via calculus and computerized mathematics. Let $a = a_1+ i a_2, b = b_1+ i b_2, b = c_1+ i c_2.\,$ We need to prove that

$\left| a_1^2+2 i a_2 a_1-a_2^2+b_1 c_1+i b_2 c_1+i b_1 c_2-b_2 c_2\right| \geq \left| b_1+i b_2+c_1+i c_2\right|$

under constraint $\sqrt{\left(a_2+b_2+c_2\right){}^2+\left(a_1+b_1+c_1\right){}^2}\leq 1.$

Since $\left| a_1+i a_2\right| =1$, etc.,

$\text{lhs}=\sqrt{\left(\left(-2 a_2^2+\sqrt{1-b_2^2} \sqrt{1-c_2^2}-b_2 c_2+1\right){}^2+\left(2 \sqrt{1-a_2^2} a_2+b_2 \sqrt{1-c_2^2}+\sqrt{1-b_2^2} c_2\right)^2\right)}$

$\text{rhs}= \sqrt{\left(\sqrt{1-b_2^2}+\sqrt{1-c_2^2}\right){}^2+\left(c_2+b_2\right){}^2}$

we need to show that $f= \text{lhs - rhs} \geq 0$ under the constraint

$\sqrt{\left(a_2+b_2+c_2\right)^2+\left(\sqrt{1-a_2^2}+\sqrt{1-b_2^2}+\sqrt{1-c_2^2}\right)^2}\leq 1.$

Now it looks that the devil is in the constraint. We find that it requires two of the variables ${a_2,b_2,c_2}$ to be at $\{1,-1\}$, stuck on the corner, while the third can take any value in $[-1,1]$.


constraint from one side

And the solution, obviously,

solution, obviously


The problem which is an invention of Leo Giugiuc and Diana Veronica, has been kindly posted by Leo Giugiuc on the CutTheKnotMath facebook page. Solution 2 is by @adisetyop; Solution 3 is by Amit Itagi and independently by @ramonaamc. The Exploration is by N. N. Taleb.


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