给定实数 $r \in(0,1)$．证明：若 $n$ 个复数 $z_{1}, z_{2}, \cdots, z_{n}$ 满足 $\left|z_{k}-1\right| \leqslant r(k=1,2, \cdots, n)$，则 $\left|z_{1}+z_{2}+\cdots+z_{n}\right|\left|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}}\right|\geqslant n^{2}\left(1-r^{2}\right)$．
【难度】
【出处】
2014第30届CMO试题
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【答案】
略
【解析】
设 $z_{k}=x_{k}+y_{k}$，其中 $x_{k}, y_{k} \in \mathbf{R}, k=1,2, \cdots, n$
先证明：$\dfrac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}} \geqslant 1-r^{2}(k=1,2, \cdots, n)$ ①
记 $u=\dfrac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}}$．
由 $\left|x_{k}-1\right| \leqslant r<1$，知 $x_{k}>0$．
则 $u>0$，且 $y_{k}^{2}=\left(\dfrac{1}{u}-1\right) x_{k}^{2}$．
故 $r^{2} \geqslant\left|z_{k}-1\right|^{2}=\left(x_{k}-1\right)^{2}+\left(\dfrac{1}{u}-1\right) x_{k}^{2}=\dfrac{1}{u}\left(x_{k}-u\right)^{2}+1-u \geqslant 1-u$．
从而，$u\geqslant 1-r^2$，即式 ① 成立．
注意到，$\displaystyle \left|z_{1}+z_{2}+\cdots+z_{n}\right|\geqslant\left|\operatorname{Re}\left(z_{1}+z_{2}+\cdots+z_{n}\right)\right|=\sum\limits_{k=1}^{n} x_{k}$．
又 $\dfrac{1}{z_{k}}=\dfrac{x_{k}-y_{k}}{x_{k}^{2}+y_{k}^{2}}(k=1,2, \cdots, n)$．故 $\displaystyle \left|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}} \geqslant\right| \operatorname{Re}\left(\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}}\right) |=\sum\limits_{k=1}^{n} \dfrac{x_{k}}{x_{k}^{2}+y_{k}^{2}}$
因为 $x_{k}>0(h=1,2, \cdots, n)$，所以，由柯西不等式得
$\left|z_{1}+z_{2}+\cdots+z_{n}\right| \dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}} |$
$\displaystyle \geqslant\left(\sum\limits_{k=1}^{n} x_{k}\right)\left(\sum_{k=1}^{n} \dfrac{x_{k}}{x_{k}^{2}+y_{k}^{2}}\right)$
$\displaystyle \geqslant\left(\sum\limits_{k=1}^{n} \sqrt{\dfrac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}}}\right) \geqslant\left(n \sqrt{1-r^{2}}\right)^{2}$
$=n^{2}\left(1-r^{2}\right)$
先证明：$\dfrac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}} \geqslant 1-r^{2}(k=1,2, \cdots, n)$ ①
记 $u=\dfrac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}}$．
由 $\left|x_{k}-1\right| \leqslant r<1$，知 $x_{k}>0$．
则 $u>0$，且 $y_{k}^{2}=\left(\dfrac{1}{u}-1\right) x_{k}^{2}$．
故 $r^{2} \geqslant\left|z_{k}-1\right|^{2}=\left(x_{k}-1\right)^{2}+\left(\dfrac{1}{u}-1\right) x_{k}^{2}=\dfrac{1}{u}\left(x_{k}-u\right)^{2}+1-u \geqslant 1-u$．
从而，$u\geqslant 1-r^2$，即式 ① 成立．
注意到，$\displaystyle \left|z_{1}+z_{2}+\cdots+z_{n}\right|\geqslant\left|\operatorname{Re}\left(z_{1}+z_{2}+\cdots+z_{n}\right)\right|=\sum\limits_{k=1}^{n} x_{k}$．
又 $\dfrac{1}{z_{k}}=\dfrac{x_{k}-y_{k}}{x_{k}^{2}+y_{k}^{2}}(k=1,2, \cdots, n)$．故 $\displaystyle \left|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}} \geqslant\right| \operatorname{Re}\left(\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}}\right) |=\sum\limits_{k=1}^{n} \dfrac{x_{k}}{x_{k}^{2}+y_{k}^{2}}$
因为 $x_{k}>0(h=1,2, \cdots, n)$，所以，由柯西不等式得
$\left|z_{1}+z_{2}+\cdots+z_{n}\right| \dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}+\cdots+\dfrac{1}{z_{n}} |$
$\displaystyle \geqslant\left(\sum\limits_{k=1}^{n} x_{k}\right)\left(\sum_{k=1}^{n} \dfrac{x_{k}}{x_{k}^{2}+y_{k}^{2}}\right)$
$\displaystyle \geqslant\left(\sum\limits_{k=1}^{n} \sqrt{\dfrac{x_{k}^{2}}{x_{k}^{2}+y_{k}^{2}}}\right) \geqslant\left(n \sqrt{1-r^{2}}\right)^{2}$
$=n^{2}\left(1-r^{2}\right)$
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