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Let $z_{i}(i=1,2,\cdots,n),n\ge 3$ be complex numbers, and 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 $(1)$?

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  • $\begingroup$ I don't think the inequality you have at the bottom holds. Consider $n=2, z_1 = 1, z_2 = 0$. $\endgroup$ – mathworker21 Mar 4 '18 at 4:40
  • $\begingroup$ In any case, it is not at all related with the original inequality, as the quotient of the sums is not the sum of the quotients. $\endgroup$ – Martin Argerami Mar 4 '18 at 4:44
  • $\begingroup$ @mathworker21, $n\ge 3$.... $\endgroup$ – function sug Mar 4 '18 at 4:50
  • $\begingroup$ @functionsug: $n=3$, $z_1=1$, $z_2=z_3=0$. $\endgroup$ – Martin Argerami Mar 4 '18 at 4:52
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    $\begingroup$ @MartinArgerami I wouldn't be so assertive. The second inequality (when it holds) implies the first one. Still, you are right: the second inequality can fail very easily, so this implication is rather useless. I fancy I can prove (1) for $n\ge 4$, but $n=3$ looks harder. $\endgroup$ – fedja Mar 29 '18 at 22:03
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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 $k$.

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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 $0<t<\frac 15$.

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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 $z_{n+1}=z_1$.

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    $\begingroup$ Isn't what you suggest as “sufficient to prove” a stronger inequality? Why should that be easier? $\endgroup$ – Martin R Mar 29 '18 at 20:44
  • $\begingroup$ @MartinR yes it is a stronger inequality that is why it is sufficient. I am not claiming it is easier, but it could be. I don't know for sure, this is just a suggestion. $\endgroup$ – Arian Mar 29 '18 at 21:02

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