# 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 $1.$

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.

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:

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

i.e.,

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

and, finally,

(1)

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

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

(2)

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

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
• $u\ge\pi,\,u+v\le 3\pi.$

### 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

$a=1,~b=e^{i\theta},~c=e^{i\phi}.$

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}

### Exploration

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 $[-1,1]$.

And the solution, obviously,

### Acknowledgment

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.