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.
 A: 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*}
A: 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
