已知函数
\(f(x)=\frac{\ln x}{x}\)，若
\(f(x)=a\)有两个不同的零点
\(x_1\)，
\(x_2\)，且
\(x_1 < x_2\)，求证：
(1) 求
\(a\)的取值范围.
(2) 
\(x_1 + x_2>\frac{2}{a}\)
(3) 
\(x_1 + x_2>2e\)
(4) 
\(\sqrt{x_1}+\sqrt{x_2}>\frac{2}{\sqrt{a}}\)
(5) 
\(\frac{1}{x_1}+\frac{1}{x_2}>2a\)
(6) 
\(e^2 < x_1x_2\)
(7) 
\(x_1x_2<\frac{1}{a^2}\)
(8) 
\(\frac{3}{a}-e<x_1 + x_2\)
(9) 
\(x_1 + x_2<\frac{-2\ln a}{a}\)
(10) 
\(x_1 + x_2>\frac{1 - \ln a}{a}\)
(11) 
\(\ln x_1+\ln x_2>1 - \ln a\)
(12) 
\(x_1x_2>\frac{e}{a}\)
(13) 
\(\frac{1}{x_1}+\frac{1}{x_2}>\frac{2}{e}\)
(14) 
\(2\ln x_1+\ln x_2>e\)
(15) 
\(\frac{1}{\ln x_1}+\frac{1}{\ln x_2}>2ae\)
(16) 
\((x_1 + 1)(x_2 + 1)<\frac{3}{a^2}-\frac{2}{a}+1\)
(17) 
\(x_1^2x_2 + x_2^2x_1>\frac{2e}{a^2}\)
(18) 
\(\frac{x_1}{x_2}<ae\)
(19) 
\(x_1<\frac{1 - \sqrt{1 - ae}}{a}\)
(20) 
\(x_2>\frac{1 + \sqrt{1 - ae}}{a}\)
(21) 
\(x_2 - x_1>\frac{2\sqrt{1 - ae}}{a}\)
(22) 
\(x_2 - x_1>\frac{\sqrt{1 - ae}}{a}\)
(23) 
\(x_2 - x_1>\sqrt{1 - ae}\)
(24) 
\(x_2 - x_1>2\sqrt{\frac{e}{a}-e^2}\)
(25) 
\(f^\prime(x_1)+f^\prime(x_2)>0\)
(26) 
\(x_2 - x_1>(e^2 - 2)(1 - ae)\)
(27) 
\(x_2 - x_1>2(e - 2)\sqrt{1 - ae}\)
(28) 
\(x_2 - x_1<\frac{4e^{\frac{3}{2}}-1-(2e^{\frac{3}{2}}+1)a}{2e^{\frac{3}{2}}}(-\frac{3}{4e^{\frac{3}{2}}}<a<\frac{1}{e})\)
(29) 证明：当
\(m\geq1\)时，
\(x_1x_2^m>e^{m + 1}\)
(30) 若
\(\frac{1}{a}<(1 - m)x_1+mx_2\)恒成立，求
\(m\)的取值范围.
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