设 ${{x}_{1}},{{x}_{2}},\cdots,{{x}_{n}}\in (0,1)$，$n\geqslant 2$．求证：
$\frac{\sqrt{1-{{x}_{1}}}}{{{x}_{1}}}+\frac{\sqrt{1-{{x}_{2}}}}{{{x}_{2}}}+\cdots+\frac{\sqrt{1-{{x}_{n}}}}{{{x}_{n}}}<\frac{\sqrt{n-1}}{{{x}_{1}}{{x}_{2}}\cdots{{x}_{n}}}$．
$\frac{\sqrt{1-{{x}_{1}}}}{{{x}_{1}}}+\frac{\sqrt{1-{{x}_{2}}}}{{{x}_{2}}}+\cdots+\frac{\sqrt{1-{{x}_{n}}}}{{{x}_{n}}}<\frac{\sqrt{n-1}}{{{x}_{1}}{{x}_{2}}\cdots{{x}_{n}}}$．
【难度】
【出处】
2015第14届CGMO试题
【标注】
【答案】
略
【解析】
对 $n$ 应用数学归纳法．当 $n=2$ 时，由Cauchy不等式可得
$\frac{\sqrt{1-{{x}_{1}}}}{{{x}_{1}}}+\frac{\sqrt{1-{{x}_{2}}}}{{{x}_{2}}}=\frac{{{x}_{2}}\sqrt{1-{{x}_{1}}}+{{x}_{1}}\sqrt{1-{{x}_{2}}}}{{{x}_{1}}{{x}_{2}}}\leqslant\frac{\sqrt{1-{{x}_{1}}+x_{1}^{2}}\sqrt{1-{{x}_{2}}+x_{2}^{2}}}{{{x}_{1}}{{x}_{2}}}<\frac{1}{{{x}_{1}}{{x}_{2}}}$
当 $n\geqslant 3$ 时，由归纳假设和Cauchy不等式，得
$\begin{matrix}
\frac{\sqrt{1-{{x}_{1}}}}{{{x}_{1}}}+\frac{\sqrt{1-{{x}_{2}}}}{{{x}_{2}}}+\cdots+\frac{\sqrt{1-{{x}_{n}}}}{{{x}_{n}}}<\frac{\sqrt{n-2}}{{{x}_{1}}{{x}_{2}}\cdots{{x}_{n-1}}}+\frac{\sqrt{1-{{x}_{n}}}}{{{x}_{n}}} \\
=\frac{\sqrt{n-2}\cdot{{x}_{n}}+{{x}_{1}}{{x}_{2}}\cdots{{x}_{n-1}}\sqrt{1-{{x}_{n}}}}{{{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}}\leqslant\frac{\sqrt{n-2+{{({{x}_{1}}{{x}_{2}}\cdots{{x}_{n-1}})}^{2}}}\sqrt{1-{{x}_{n}}+x_{n}^{2}}}{{{x}_{1}}{{x}_{2}}\cdots{{x}_{n}}}<\frac{\sqrt{n-1}}{{{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}} \\
\end{matrix}$
$\frac{\sqrt{1-{{x}_{1}}}}{{{x}_{1}}}+\frac{\sqrt{1-{{x}_{2}}}}{{{x}_{2}}}=\frac{{{x}_{2}}\sqrt{1-{{x}_{1}}}+{{x}_{1}}\sqrt{1-{{x}_{2}}}}{{{x}_{1}}{{x}_{2}}}\leqslant\frac{\sqrt{1-{{x}_{1}}+x_{1}^{2}}\sqrt{1-{{x}_{2}}+x_{2}^{2}}}{{{x}_{1}}{{x}_{2}}}<\frac{1}{{{x}_{1}}{{x}_{2}}}$
当 $n\geqslant 3$ 时，由归纳假设和Cauchy不等式，得
$\begin{matrix}
\frac{\sqrt{1-{{x}_{1}}}}{{{x}_{1}}}+\frac{\sqrt{1-{{x}_{2}}}}{{{x}_{2}}}+\cdots+\frac{\sqrt{1-{{x}_{n}}}}{{{x}_{n}}}<\frac{\sqrt{n-2}}{{{x}_{1}}{{x}_{2}}\cdots{{x}_{n-1}}}+\frac{\sqrt{1-{{x}_{n}}}}{{{x}_{n}}} \\
=\frac{\sqrt{n-2}\cdot{{x}_{n}}+{{x}_{1}}{{x}_{2}}\cdots{{x}_{n-1}}\sqrt{1-{{x}_{n}}}}{{{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}}\leqslant\frac{\sqrt{n-2+{{({{x}_{1}}{{x}_{2}}\cdots{{x}_{n-1}})}^{2}}}\sqrt{1-{{x}_{n}}+x_{n}^{2}}}{{{x}_{1}}{{x}_{2}}\cdots{{x}_{n}}}<\frac{\sqrt{n-1}}{{{x}_{1}}{{x}_{2}}\cdots {{x}_{n}}} \\
\end{matrix}$
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