Let $r$ $z_{1},z_{2},\cdots,z_{n}$ be given such that $$ |z_{i}-1|\le r,i=1,2,\cdots,n,r\in(0,1). $$ Show that $$|z_{1}+z_{2}+\cdots+z_{n}|\left|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}}\right|\ge n^2(1-r^2).$$ Can we use the identity from http://en.wikipedia.org/wiki/Lagrange%27s_identity

I know that if $z_{i}\in R$, then using Cauchy-Schwarz inequality we have $$(z_{1}+z_{2}+\cdots+z_{n})\left(\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}}\right)\ge n^2. $$ This problem is from the China 2014 Mathematical Olympiad Contest.


Note that $|z_i-1|\leq r$ implies $|\arg z_i|\leq\cos^{-1}\sqrt{1-r^2}$.

We shall generalise the inequality. Fix $0<\theta<90^\circ$; we shall show $$\left|\sum z_i\right|\left|\sum\frac1{z_i}\right|\geq n^2\cos^2\theta$$ for nonzero complex numbers $z_1,\ldots,z_n$ with $|\arg z_i|\leq\theta$. This immediately implies the desired inequality.

Now $\Re(z_i)=|z_i|\cos\arg z_i$ and $\Re\left(\frac1{z_i}\right)=\frac1{|z_i|}\cos\arg z_i$, so $$\Re(z_i)\Re\left(\frac1{z_i}\right)=\cos^2\arg z_i\geq\cos^2\theta.$$ Therefore [\begin{align*} \left|\sum z_i\right|\left|\sum\frac1{z_i}\right| &\geq\Re\left(\sum z_i\right)\Re\left(\sum\frac1{z_i}\right)\quad(\because |z|\geq\Re(z))\\ &=\left(\sum\Re(z_i)\right)\left(\sum\Re\left(\frac1{z_i}\right)\right)\\ &\geq\left(\sum\sqrt{\Re(z_i)\Re\left(\frac1{z_i}\right)}\right)^2\quad\mbox{(C-S)}\\ &\geq(n\cos\theta)^2, \end{align*}

| cite | improve this answer | |
  • 2
    $\begingroup$ I wonder why this looks so familiar... $\endgroup$ – chronondecay Jan 14 '15 at 2:46

It follows from this classical generalization of the AM-GM inequality

If $z_j=\rho_je^{i\theta_j}$ where $|\theta_j|<\phi$ and $\phi<\frac{\pi}{2}$

then $\left(\cos{\phi}\right) |z_1z_2...z_n|^{1/n}\leq\frac{1}{n}|z_1+z_2+...+z_n|$

this is shown by noting that $RHS \geq \frac{1}{n}*Re(z_1+z_2+...+z_n)=\frac{1}{n}*(Re(z_1)+Re(z_2)+...+Re(z_n))$

Note $Re(z_j)\geq\rho_j\cos{\phi}$ so the lemma follows from regular AM-GM

From geometry we know that in your problem the complex numbers have arguments $|\theta| \leq \sin^{-1}{r}$

This lemma shows that the LHS (left hand side) of your inequality $\geq (1-r^2)*n^2$ as desired

| cite | improve this answer | |
  • $\begingroup$ I think your methods is not true $\endgroup$ – math110 Dec 22 '14 at 17:12
  • $\begingroup$ explain? it's very similar to the one you just put up but with AMGM $\endgroup$ – davik Dec 22 '14 at 17:15
  • $\begingroup$ this $(\cos{\phi})|z_{1}z_{2}\cdots z_{n}|^{1/n}\le\dfrac{1}{n}|z_{1}+z_{2}\cdots z_{n}|?$ this is why? $\endgroup$ – math110 Dec 22 '14 at 17:21
  • $\begingroup$ RHS $\geq \frac{1}{n}*(Re(z_1)+Re(z_2)...Re(z_n)) \geq \frac{1}{n}(\cos{\phi})(\rho_1+...\rho_n)\geq (\cos{\phi})(\rho_1\rho_2...\rho_n)^{1/n}$ $\endgroup$ – davik Dec 22 '14 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.