已知函数
[imath]f(x)=\ln x-a x(a \in \mathbb{R}),[/imath]
若
[imath]f(x)[/imath]
有两个零点
[imath]x_{1}, x_{2}\left(x_{1}<x_{2}\right), \mathrm{e}=2.71828 \cdots[/imath]
为自然对数的底数，求证:
(1)
[imath]0<a<\frac{1}{\mathrm{e}}[/imath]
(2)
[imath]x_{1}+x_{2}>\frac{2}{a}[/imath]
(3)
[imath]x_{1}+x_{2}>2 \mathrm{e} [/imath]
(4)
[imath]x_{1} x_{2}<\frac{1}{a^{2}}[/imath]
(5)
[imath]x_{1} x_{2}>\mathrm{e}^{2}[/imath]
(6)
[imath]\sqrt{x_{1}}+\sqrt{x_{2}}>\frac{2}{\sqrt{a}}[/imath]
(7)
[imath]\frac{1}{x_{1}}+\frac{1}{x_{2}}>\frac{2}{\mathrm{e}} [/imath]
(8)
[imath]\frac{1}{x_{1}}+\frac{1}{x_{2}}>2 a[/imath]
(9)
[imath]\frac{1}{\ln x_{1}}+\frac{1}{\ln x_{2}}>2 a \mathrm{e}[/imath]
(10)
[imath]x_{1}+x_{2}>\frac{1-\ln a}{a} [/imath]
(11)
[imath]\ln x_{1}+\ln x_{2}>1-\ln a [/imath]
(12)
[imath]x_{1} x_{2}>\frac{\mathrm{e}}{a}[/imath]
(13)
[imath]x_{1}+x_{2}>\frac{3}{a}-\mathrm{e}[/imath]
(14)
[imath]x_{1}^{2} x_{2}+x_{2}^{2} x_{1}>\frac{2 \mathrm{e}}{a^{2}}[/imath]
(15)
[imath]2 \ln x_{1}+\ln x_{2}>\mathrm{e}[/imath]
(16)
[imath]x_{1}+x_{2}<-\frac{2 \ln a}{a}[/imath]
(17)
[imath]\left(x_{1}+1\right)\left(x_{2}+1\right)<\frac{3}{a^{2}}-\frac{2}{a}+1[/imath]
(18)
[imath]\frac{x_{1}}{x_{2}}<a \mathrm{e}[/imath]
(19)
[imath]x_{1}<\frac{1-\sqrt{1-a \mathrm{e}}}{a} [/imath]
(20)
[imath]x_{2}>\frac{1+\sqrt{1-a \mathrm{e}}}{a} [/imath]
(21)
[imath]x_{2}-x_{1}>\frac{2 \sqrt{1-a \mathrm{e}}}{a}[/imath]
(22)
[imath]x_{2}-x_{1}>\frac{\sqrt{1-a \mathrm{e}}}{a}[/imath]
(23)
[imath]x_{2}-x_{1}>\sqrt{1-a \mathrm{e}}[/imath]
(24)
[imath]x_{2}-x_{1}>2 \sqrt{\frac{\mathrm{e}}{a}-\mathrm{e}^{2}}[/imath]
(25)
[imath]x_{2}-x_{1}>\left(\mathrm{e}^{2}-2\right)(1-a \mathrm{e})[/imath]
(26)
[imath]x_{2}-x_{1}>2(\mathrm{e}-2) \sqrt{1-a \mathrm{e}}[/imath]
(27) 当
[imath]\frac{3}{2 \mathrm{e}^{\frac{3}{2}}}<a<\frac{1}{\mathrm{e}}[/imath]
时,
[imath]x_{2}-x_{1}<4 \mathrm{e}^{\frac{3}{2}}-1-\left(2 \mathrm{e}^{2}+1\right) a[/imath]
(28) 当
[imath]m \geq 1[/imath]
时,
[imath]x_{1} x_{2}^{m}>\mathrm{e}^{m+1}[/imath]
(29) 设
[imath]g(x)=\frac{\ln x}{x},[/imath]
求证:
[imath]g^{\prime}\left(x_{1}\right)+g^{\prime}\left(x_{2}\right)>0[/imath]
.
(30) 若
[imath]\frac{1}{a}<(1-m) x_{1}+m x_{2}[/imath]
恒成立，求
[imath]m[/imath]
的取值范围.
