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本帖最后由 hbghlyj 于 2022-10-24 20:54 编辑
记\[f(x) = x^{x},g(x) = \frac{1}{f\left( \frac{1}{x + 1} \right)} + f\left( \frac{1}{x} \right) - 2\]则\[g^{'}(x) = \frac{1 -\ln(1 + x)}{(x + 1)^{2}f\left( \frac{1}{x + 1} \right)} - \frac{f\left( \frac{1}{x} \right)\left( 1 - \ln x \right)}{x^{2}}\]显然,当$e-1≤x≤e$时,有$g'(x)<0$当$x<e-1$时,先证明\[\frac{1}{(x + 1)^{2}f\left( \frac{1}{x + 1} \right)} - \frac{f\left( \frac{1}{x} \right)}{x^{2}} < 0\]这等价于\[\frac{x^{2}}{(x + 1)^{2}} - f\left( \frac{1}{x} \right)f\left( \frac{1}{x + 1} \right) < 0\]即\[\frac{\ln x}{x} + \frac{\ln{(x + 1)}}{x + 1} + 2\ln{\left( 1 - \frac{1}{x + 1} \right) < 0}\]记\[h(x) = \left( 1 + \frac{1}{x} \right)\ln x + \ln{(x + 1)} - 2\]则\[h^{'}(x) = \frac{1}{x + 1} + \frac{x + 1 - \ln x}{x^{2}} > 0\]故\[h(x) < h(e - 1) = \frac{e}{e - 1}\ln(e - 1) - 1 < 0\]因此\[\frac{\ln x}{x} + \frac{\ln{(x + 1)}}{x + 1} + 2\ln{\left( 1 - \frac{1}{x + 1} \right) < \frac{\ln x}{x} + \frac{\ln{(x + 1)}}{x + 1} - \frac{2}{x + 1} = \frac{h(x)}{x + 1} < 0}\]故\[g^{'}(x) < \left\lbrack 1 - \ln{(x + 1)} \right\rbrack\left\lbrack \frac{1}{(x + 1)^{2}f\left( \frac{1}{x + 1} \right)} - \frac{f\left( \frac{1}{x} \right)}{x^{2}} \right\rbrack < 0,x < e - 1\]当$x>e$时,先证明\[\frac{1}{(x + 1)^{2}f\left( \frac{1}{x + 1} \right)} - \frac{f\left( \frac{1}{x} \right)}{x^{2}} > 0\]这等价于\[\frac{x^{2}}{(x + 1)^{2}} - f\left( \frac{1}{x} \right)f\left( \frac{1}{x + 1} \right) > 0\]即\[\frac{\ln x}{x} + \frac{\ln{(x + 1)}}{x + 1} + 2\ln{\left( 1 - \frac{1}{x + 1} \right) > 0}\]记\[i(x) = \frac{1}{x} + \frac{1}{x + 1} + 2\ln\left( 1 - \frac{1}{x + 1} \right)\]则\[i^{'}(x) = - \frac{1}{x^{2}(x + 1)^{2}} < 0\]故\[i(x) > \lim_{x \rightarrow \infty}{i(x) = 0}\]于是\[\frac{\ln x}{x} + \frac{\ln{(x + 1)}}{x + 1} + 2\ln{\left( 1 - \frac{1}{x + 1} \right) > i(x) > 0}\]故\[g^{'}(x) < \left( 1 - \ln x \right)\left\lbrack \frac{1}{(x + 1)^{2}f\left( \frac{1}{x + 1} \right)} - \frac{f\left( \frac{1}{x} \right)}{x^{2}} \right\rbrack < 0,x > e\]综上:$g'(x)<0$.即$g(x)$单调递减.因此\[g(x) > \lim_{x \rightarrow \infty}{g(x) = 0}\]即\[(1 + x)^{\frac{1}{1 + x}} + x^{- \frac{1}{x}} > 2\] |
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