Prove that $\sum\limits_{k=1}^{n}\frac{|z_{k}|^2}{|z_{k+1}-z_{k}|^2}\ge 1$ for complex numbers $z_k$'s

Question

Let $n \geq 3$. Let $z_{i}$ (for $i=1,2,\cdots,n$) be complex numbers. Assume that $z_{k+1} - z_k \ne 0$ for $k=1,2, \cdots, n-1$ and $z_n - z_1 \ne 0$. Show that

$$\sum_{k=1}^{n}\dfrac{|z_{k}|^2}{|z_{k+1}-z_{k}|^2}\ge 1\tag{1},$$

where $z_{n+1}=z_{1}$.

I think can use the Cauchy-Schwarz inequality to solve it. But I think this complex inequality can't hold:

$$\left(\sum_{k=1}^{n}|z_{k}|\right)^2\ge \sum_{k=1}^{n}|z_{k+1}-z_{k}|^2.$$

So how do I prove $(1)$?

Answer 1

Not a proof, but here are some ideas.

Note that if $z_k=0$ for any $k$, the inequality holds trivially. So we may assume $z_k\ne0$ for all $k$. Then, $$\tag1 \sum_{k=1}^{n}\dfrac{|z_{k}|^2}{|z_{k+1}-z_{k}|^2} =\sum_{k=1}^{n}\dfrac{1}{|\frac{z_{k+1}}{z_{k}}-1|^2}. $$ Thus the inequality may be rewritten as $$\tag2 \sum_{k=1}^{n}\dfrac{1}{|\alpha_{k}-1|^2}\ge 1, $$ where $\alpha_1\cdots\alpha_n=1$.

If $|\alpha_k-1|<1$ for any $k$, then $(2)$ holds trivially. So we may further assume that $|\alpha_k-1|\geq1$ for all $k$.

Answer 2

Too long for a comment. First any two consecutive terms must be different otherwise it leads to division by zero. Next $$\sum_{k=1}^n\frac{|z_k|^2}{|z_{k+1}-z_k|^2}=\sum_{k=1}^n\frac{|z_{k+1}-(z_{k+1}-z_k)|^2}{|z_{k+1}-z_k|^2}\geqslant\sum_{k=1}^n\frac{||z_{k+1}|-|z_{k+1}-z_k||^2}{|z_{k+1}-z_k|^2}$$ by triangle inequality. The last sum is equivalent to $$\sum_{k=1}^n\frac{||z_{k+1}|-|z_{k+1}-z_k||^2}{|z_{k+1}-z_k|^2}=\sum_{k=1}^n\Big(\frac{|z_{k+1}|}{|z_{k+1}-z_k|}-1\Big)^2$$ Therefore it is sufficient to prove the inequality $$\sum_{k=1}^n\Big(\frac{|z_{k+1}|}{|z_{k+1}-z_k|}-1\Big)^2\geqslant 1$$ If this is not the case then for all $k$ we must have $|z_{k+1}|<2|z_{k+1}-z_k|$. Maybe this could lead to some contradictions using also $z_{n+1}=z_1$.

Answer 3

Let's try to do $n\ge 4$ at least. Let $x_k=|z_k|$. Then it will suffice to show that $$ \Phi(x)=\sum_k\frac{x_k^2}{(x_k+x_{k+1})^2}\ge 1 $$ Let us consider $\Phi$ on $x_k>0,\sum_k x_k=1$.

Notice first of all that if $\Phi(x)\le m<1$, then $x$ is separated from the boundary in the sense that $x_k\ge c(m)>0$ for all $k$. Indeed, at least one $x_k\ge 1/n$. Since $m<1$, we must have $x_{k+1}\ge\lambda(m)x_k$ (otherwise one single term is already above $m$. Then $x_{k+2}\ge\lambda(m)x_{k+1}$ and so on over the cycle, so all $x_j\ge n^{-1}\lambda(m)^{n-1}$. Thus, only the critical points are of interest. Also, since the functional is homogeneous of degree $0$, the differential should vanish at the critical points on the whole space, not just along the hyperplane $\sum_k dx_k=0$.

Now, $$ x_{k}\frac{\partial \Phi(x)}{\partial x_k}=2\left[-\frac{x_{k-1}^2x_k}{(x_{k-1}+x_{k})^3}+\frac{x_{k}^2x_{k+1}}{(x_{k}+x_{k+1})^3}\right] $$ which means that at any critical point all ratios $w_k=\frac{x_{k+1}}{x_k}$ are roots of $(1+w)^3=sw$ with some common $s>0$.

Now comes some casework. Notice that this equation can have only two positive roots (the LHS is strictly convex). When $s=8$, $w=1$ is a root and it is the larger one (because the tangent line to $(1+w)^3$ at $w=1$ has slope $12>8$. Thus, when $s<8$, both roots are below $1$, which makes this case impossible because then $x_k$ would decrease over the cycle. When $s=8$, the only option is to have all $w_k=1$, which results in $\Phi(x)=\frac n4\ge 1$ as long as $n\ge 4$. So the only interesting case is $s>8$ when there are two roots $y<1<z$. Notice that we cannot use $z$ every time (there are no strictly increasing cycles). If we use $y$ at least once, we have $$ \Phi(x)\ge \frac{1}{(1+y)^2}+\frac{n-1}{(1+z)^2} = \frac{1}{(1+y)^2}+\frac{(n-1)(1+z)}{(1+z)^3} \\ \ge \frac{1}{(1+y)^2}+\frac{3z}{(1+z)^3}=\frac{1}{(1+y)^2}+\frac 3s\,. $$
If we show that the last expression is above $1$, we are done.

To this end, it will suffice to show that $(1+y)^3<sy$ for $y=\sqrt{\frac s{s-3}}-1$ (the value that makes the expression we are interested in exactly $1$. This rewrites as $$ s\left(\sqrt{\frac s{s-3}}-1\right)>\sqrt{\frac s{s-3}}^3\,, $$ then $$ \left(s-\frac s{s-3}\right)\sqrt{\frac s{s-3}}>s $$ and, denoting $t=\frac 1{s-3}<\frac 15$, $$ (1-t)\sqrt{1+3t}> 1\,, $$ or, finally, $$ (1-t)^2(1+3t)=1+t-5t^2+3t^3> 1\,, $$ which is obvious for $0<t<\frac 15$.