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Last edited by realnumber 2014-4-25 12:23当$x=\frac{1}{e}$时，显然成立．
当$x>\frac{1}{e}$时，$1+x+x\ln{x}+\frac{1}{1+\ln{x}}\ge p$．
当$0<x＜\frac{1}{e}$时，$1+x+x\ln{x}+\frac{1}{1+\ln{x}}\le p$．其中$1+x+x\ln{x}+\frac{1}{1+\ln{x}}＜１$显然．
lm1.gsp
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$1\le p\le x_0$,$x_0$约为2.904.....． |
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