张祖锦第7卷第483期一个对数-平方根不等式

设 $n$ 为正整数, $0<y_i\leq x_i<1$, $1\leq i\leq n$. 试证:

$$\bex \frac{\ln x_1+\cdots +\ln x_n}{\ln y_1+\cdots +\ln y_n} \leq \sqrt{\frac{1-x_1}{1-y_1}+\cdots +\frac{1-x_n}{1-y_n}}. \eex$$

证明: 当 $n=1$ 时,

$$\beex \bea &\quad \frac{\ln x_1}{\ln y_1}\leq \sqrt{\frac{1-x_1}{1-y_1}}\\ &\la \frac{\ln x_1}{\sqrt{1-x_1}}\geq \frac{\ln y_1}{\sqrt{1-y_1}}\\ &\la f(x)= \frac{\ln x}{\sqrt{1-x}}\mbox{ 是 }(0,1)\mbox{ 上的递增函数}\\ &\la f'(x)=\frac{2-2x+x\ln x}{2x(1-x)^\frac{3}{2}}\geq 0\\ &\la g(x)= 2-2x+x\ln x\geq 0\\ &\la\seddm{ g'(x)=\ln x-1\leq 0\quad\sex{0<x<1}\\ g(1)=0}. \eea \eeex$$ 当 $n\geq 2$ 时, 由 $n=1$ 时的结论, $$\bex \frac{\ln x_1+\cdots +\ln x_n}{\ln y_1+\cdots +\ln y_n} =\frac{\ln (x_1\cdots x_n)}{\ln(y_1\cdots y_n)} \leq \sqrt{\frac{1-x_1\cdots x_n}{1-y_1\cdots y_n}}. \eex$$ 而仅需证明 $$\bex \frac{1-x_1\cdots x_n}{1-y_1\cdots y_n} \leq \frac{1-x_1}{1-y_1} +\cdots +\frac{1-x_n}{1-y_n}. \eex$$ 其可用数学归纳法证明. 当 $n=1$ 时, 结论自明. 当 $n=2$ 时, $$\beex \bea \frac{1-x_1x_2}{1-y_1y_2} &<\frac{(1-x_1)+(1-x_2)}{1-y_1y_2} =\frac{1-x_1}{1-y_1y_2}+\frac{1-x_2}{1-y_1y_2}\\ &<\frac{1-x_1}{1-y_1}+\frac{1-x_2}{1-y_2}. \eea \eeex$$ 假设当 $n=k$ 时结论成立, 则当 $n=k+1$ 时, $$\beex \bea \frac{1-x_1\cdots x_{k+1}}{1-y_1\cdots y_{k+1}} &<\frac{1-x_1}{1-y_1}+ \cdots +\frac{1-x_{k-1}}{1-y_{k-1}} +\frac{1-x_kx_{k+1}}{1-y_ky_{k+1}}\\ &\quad\sex{n=k \mbox{ 时的结论}}\\ &<\frac{1-x_1}{1-y_1}+\cdots +\frac{1-x_{k+1}}{1-y_{k+1}}\\ &\quad\sex{n=2 \mbox{ 时的结论}}. \eea \eeex$$

 

杂志目录见: http://www.cnblogs.com/zhangzujin/p/3527416.html