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.