# How prove this complex inequality with same as (2014 china CMO) Cauchy-Schwarz inequality

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*}

• I wonder why this looks so familiar... – 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

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