设 $n$ 为正整数, $0<y_ileq x_i<1$, $1leq ileq n$. 试证: $$ex frac{ln x_1+cdots +ln x_n}{ln y_1+cdots +ln y_n} leq sqrt{frac{1-x_1}{1-y_1}+cdots +frac{1-x_n}{1-y_n}}. eex$$
证明: 当 $n=1$ 时, $$eex ea &quad frac{ln x_1}{ln y_1}leq sqrt{frac{1-x_1}{1-y_1}}\ &la frac{ln x_1}{sqrt{1-x_1}}geq frac{ln y_1}{sqrt{1-y_1}}\ &la f(x)= frac{ln x}{sqrt{1-x}}mbox{ 是 }(0,1)mbox{ 上的递增函数}\ &la f'(x)=frac{2-2x+xln x}{2x(1-x)^frac{3}{2}}geq 0\ &la g(x)= 2-2x+xln xgeq 0\ &laseddm{ g'(x)=ln x-1leq 0quadsex{0<x<1}\ g(1)=0}. eea eeex$$ 当 $ngeq 2$ 时, 由 $n=1$ 时的结论, $$ex frac{ln x_1+cdots +ln x_n}{ln y_1+cdots +ln y_n} =frac{ln (x_1cdots x_n)}{ln(y_1cdots y_n)} leq sqrt{frac{1-x_1cdots x_n}{1-y_1cdots y_n}}. eex$$ 而仅需证明 $$ex frac{1-x_1cdots x_n}{1-y_1cdots y_n} leq frac{1-x_1}{1-y_1} +cdots +frac{1-x_n}{1-y_n}. eex$$ 其可用数学归纳法证明. 当 $n=1$ 时, 结论自明. 当 $n=2$ 时, $$eex ea frac{1-x_1x_2}{1-y_1y_2} &<frac{(1-x_1)+(1-x_2)}{1-y_1y_2} =frac{1-x_1}{1-y_1y_2}+frac{1-x_2}{1-y_1y_2}\ &<frac{1-x_1}{1-y_1}+frac{1-x_2}{1-y_2}. eea eeex$$ 假设当 $n=k$ 时结论成立, 则当 $n=k+1$ 时, $$eex ea frac{1-x_1cdots x_{k+1}}{1-y_1cdots y_{k+1}} &<frac{1-x_1}{1-y_1}+ cdots +frac{1-x_{k-1}}{1-y_{k-1}} +frac{1-x_kx_{k+1}}{1-y_ky_{k+1}}\ &quadsex{n=k mbox{ 时的结论}}\ &<frac{1-x_1}{1-y_1}+cdots +frac{1-x_{k+1}}{1-y_{k+1}}\ &quadsex{n=2 mbox{ 时的结论}}. eea eeex$$