I got stuck at the following problem. Prove that for every three distinct complex numbers $a$, $b$, $c$ with $|a| = |b| = |c| > 0$, the following inequality holds:

$ \sum_{\text{cyclic}} |(a+b)(b-c)(c-a)|^2\ \geq \left|\sum_{\text{cyclic}} (a-b)(b+c)(c+a)\right|^2 $

If we divide the inequality by $|(a+b)(b+c)(c+a)|^2$, the inequality reduces to:

$ \sum_{\text{cyclic}} \left|\frac{b-c}{b+c}\right|^2 \cdot \left|\frac{c-a}{c+a}\right|^2 \geq \left|\sum_{\text{cyclic}} \left(\frac{a-b}{a+b}\right)\right|^2 $

Then we write $a=r(\cos(t_1)+i\sin(t_1))$, $b=r(\cos(t_2)+i\sin(t_2))$, and $c=r(\cos(t_3)+i\sin(t_3))$. Thus,

$ \frac{a-b}{a+b} = i\tan\left(\frac{t_1-t_2}{2}\right) $

So the inequality we have to prove becomes:

$ \sum_{\text{cyclic}} \left(\tan\left(\frac{t_1-t_2}{2}\right)\right)^2 \cdot \left(\tan\left(\frac{t_1-t_3}{2}\right)\right)^2 \geq \left(\sum_{\text{cyclic}} \tan\left(\frac{t_2-t_3}{2}\right)\right)^2 $

Then I tried denoting $x = \frac{t_1 - t_2}{2}$, $y = \frac{t_2 - t_3}{2}$, $z = \frac{t_3 - t_1}{2}$, where x+y+z=0. Thus, we have to prove that:

$ \sum_{\text{cyclic}} (\tan(x))^2 \cdot (\tan(y))^2 \geq (\sum_{\text{cyclic}} \tan(z))^2 $ and I got stuck here. How can I continue the problem?


1 Answer 1


This solution is almost surely not the best, but it works.

$\color{#FF6200}{\bf{Lemma.}}$ If $\lambda\in\mathbb{R}$, then $\forall X\geq4\lambda:Q_\lambda(X):=X^2-\lambda(2+\lambda)X+\lambda^2(1-\lambda)^2\geq0$.

Proof. Note that the discriminant is $\Delta=\lambda^2(2+\lambda)^2-4\lambda^2(1-\lambda)^2=3\lambda^3(4-\lambda)$. So, if $\lambda\not\in[0,4]$, then $\Delta<0$ which means that it has no real solutions and, since the polynomial is $\geq0$ at $X=0$, it follows that it is $\geq0$ everywhere (in particular for $X\geq4\lambda$). So let's study the case $\lambda\in[0,4]$. Note that $$Q_\lambda(4\lambda)=16\lambda^2+\lambda(2+\lambda)4\lambda+\lambda^2(1-\lambda)^2=\lambda^4-6\lambda^3+9\lambda^2=\lambda^2(\lambda-3)^2\geq0$$ Moreover, $4\lambda\geq\lambda\cdot3\geq\frac{1}{2}\lambda(2+\lambda)$ where $X=\frac{1}{2}\lambda(2+\lambda)$ is the vertex of the convex parabola. But this means that $\forall X\geq 4\lambda$ the function $Q_\lambda(X)$ is increasing and, thus, $Q_\lambda(X)\geq Q_\lambda(4\lambda)\geq0$. $\ \square$

$\color{#FF3200}{\bf{Proposition.}}$ If $x+y+z=0$ are real numbers not in $\frac{\pi}{2}+\pi\mathbb{Z}$, then $$\sum_\text{cyc}\tan^2(x)\tan^2(y)\geq\left(\sum_\text{cyc}\tan(z)\right)^2$$

Proof. From now on let's denote $P:=\tan(x)\tan(y)$ and $S:=\tan(x)+\tan(y)$. Note that $$z=-(x+y)\Rightarrow\tan(z)=-\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}=-\frac{S}{1-P}$$ Now observe that, by the previous lemma, since $S^2\geq 4P$ by the AM-GM inequality $$S^4-P(2+P)S^2+P^2(1-P)^2\geq0$$ $$\Rightarrow S^4-2PS^2+P^2(1-P)^2\geq P^2S^2$$ $$\Rightarrow S^2(S^2-2P)+P^2(1-P)^2\geq P^2S^2$$ Now, if $P=1$ we would then have $$\tan(x)=\frac{1}{\tan(y)}=\tan\left(\frac{\pi}{2}-y\right)\Rightarrow z=-(x+y)\in\frac{\pi}{2}+\pi\mathbb{Z}$$ which contradicts the hypothesis. Hence, we can divide by $(1-P)^2$ on the previous inequality $$\frac{S^2}{(1-P)^2}(S^2-2P)+P^2\geq \frac{P^2S^2}{(1-P)^2}$$ $$\Rightarrow P^2+(S^2-2P)\left(\frac{-S}{1-P}\right)^2\geq\left(S-\frac{S}{1-P}\right)^2$$ $$\Rightarrow \tan^2(x)\tan^2(y)+(\tan^2(x)+\tan^2(y))\tan^2(z)\geq\left(\tan(x)+\tan(y)+\tan(z)\right)^2$$ $$\therefore\sum_\text{cyc}\tan^2(x)\tan^2(y)\geq\left(\sum_\text{cyc}\tan(z)\right)^2$$

As an observation, note that we have to assume that neither of $x,y,z$ are in $\frac{\pi}{2}+\pi\mathbb{Z}$ because, otherwise, the tangent isn't defined. These degenerate cases come from the step of dividing by $|\prod\limits_\text{cyc}(a+b)|^2$ where you have to assume that it is $\neq0$. But this is not too big of a deal since the case $a+b=0$ is quite easy to handle (and the other degenerate cases follow directly from this one by symmetry).


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