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[不等式] 早起一发 指对混合不等式

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v6mm131 Posted 2017-7-21 07:28 |Read mode
\[e^x-e^2\ln x>1\]

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 Author| v6mm131 Posted 2017-7-21 07:32
待定函数一发\[e^x-e^{1.5}x\ge-\frac{1}{2}e^{1.5}>1-\frac{e^2}{2}\ge e^2\ln x-e^{1.5}x+1\]

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