已知函数\(f\left ( {x} \right )=\ln {x}-ax\)有两个零点\({x}_{1}<{x}_{2}\)，求证：\({x}_{1}{x}_{2}>{e}^{2}\).

法一：分析构造法

由\(f\left ( {x} \right )=\ln {x}-ax⇔a=\frac {\ln {x}} {x}\)，令\(g\left ( {x} \right )=\frac {\ln {x}} {x}-a\)，有\(g\left ( {{x}_{1}} \right )=g\left ( {{x}_{2}} \right )=0\)，\({g}^{\, '}\left ( {x} \right )=\frac {1-\ln {x}} {{x}^{2}}\)，则\(g\left ( {x} \right )\)在\(\left ( {0，e} \right )\)上递增，\(\left ( {e，+\infty } \right )\)上递减

不妨设\(1<{x}_{1}<e<{x}_{2}\)

要证： \({x_1}{x_2} > {e^2}\) ，只需证：\({x}_{1}>\frac {{e}^{2}} {{x}_{2}}\)，只需证：\(g\left ( {{x}_{1}} \right )>g\left ( {\frac {{e}^{2}} {{x}_{2}}} \right )\)，只需证： \(g\left ( {{x}_{2}} \right )>g\left ( {\frac {{e}^{2}} {{x}_{2}}} \right )\) ，

\(F\left ( {x} \right )=g\left ( {x} \right )-g\left ( {\frac {{e}^{2}} {x}} \right )\left ( {x>e} \right )\) ， \({F}^{\, '}\left ( {x} \right )={g}^{'}\left ( {x} \right )-{\frac {{e}^{2}} {{x}^{2}}g}^{,}\left ( {\frac {{e}^{2}} {x}} \right )=\frac {\left ( {\ln {x-1}} \right )\left ( {{x}^{2}-{e}^{2}} \right )} {{e}^{2}{x}^{2}}\)

则\(F\left ( {x} \right )>F\left ( {e} \right )=0\)，即证。

法二：构造一元偏差函数

要证 \({x_1}{x_2} > {{\rm{e}}^2} \Leftrightarrow \ln \left( {{x_1}{x_2}} \right) > 2 \Leftrightarrow \ln {x_1} + \ln {x_2} > 2\) ，由 \(\ln {x_1} = a{x_1}\) ， \(\ln {x_2} = a{x_2},\) 只需证\({x}_{1}+{x}_{2}>\frac {2} {a}\)

构造函数\(F\left ( {x} \right )=f\left ( {\frac {1} {a}+x} \right )-f\left ( {\frac {1} {a}-x} \right )\left ( {x>\frac {1} {a}} \right )\)，易证。

法三：对数均值不等式

因为 \(f\left( x \right)\) 有两个相异的零点，又由于 \(x > 0\) ，故不妨设令 \({x_1} > {x_2} > 0\) ，

且有 \(\ln {x_1} = a{x_1}\) ， \(\ln {x_2} = a{x_2},\)

 \(\therefore \ln {x_1} + \ln {x_2} = a\left( {{x_1} + {x_2}} \right),\ln {x_1} - \ln {x_2} = a\left( {{x_1} - {x_2}} \right)\) ，

要证 \({x_1}{x_2} > {{\rm{e}}^2} \Leftrightarrow \ln \left( {{x_1}{x_2}} \right) > 2 \Leftrightarrow \ln {x_1} + \ln {x_2} > 2\)

\( \Leftrightarrow （{x_1} + {x_2}）\frac{{\ln {x_1} - \ln {x_2}}}{{{x_1} - {x_2}}} > 2 \Leftrightarrow \frac{{\ln {x_1} - \ln {x_2}}}{{{x_1} - {x_2}}} > \frac{2}{{{x_1} + {x_2}}}\)

\( \Leftrightarrow \ln {x_1} - \ln {x_2} > \frac{{2\left( {{x_1} - {x_2}} \right)}}{{{x_1} + {x_2}}} \Leftrightarrow \ln \frac{{{x_1}}}{{{x_2}}} > \frac{{2\left( {\frac{{{x_1}}}{{{x_2}}} - 1} \right)}}{{\frac{{{x_1}}}{{{x_2}}} + 1}}\)

令 \(t = \frac{{{x_1}}}{{{x_2}}}\) ，则 \(t > 1\) ，所以只要证明 \(\ln t > \frac{{2\left( {t - 1} \right)}}{{t + 1}},t > 1\) 时恒成立，

令 \(g\left( t \right) = \ln t - \frac{{2\left( {t - 1} \right)}}{{t + 1}}\)，\(t > 1\)

\({\rm{g'}}\left( t \right) = \frac{1}{t} - \frac{4}{{{{\left( {t + 1} \right)}^2}}} = \frac{{{{\left( {t - 1} \right)}^2}}}{{t{{\left( {t + 1} \right)}^2}}},\) 由于已知 \(t > 1\therefore g'\left( t \right) > 0\) 恒成立，

所以 \(g\left( t \right)\) 在 \(（1, + \infty ）\) 递增， \(\therefore g\left( t \right) > g\left( 1 \right) = 0\)

所以 \(t > 1\) 时， \(g\left( t \right) > 0\) 恒成立，即 \(\ln t > \frac{{2\left( {t - 1} \right)}}{{t + 1}}\) 恒成立，从而证明 \({x_1}{x_2} > {e^2}\).

法四：比值代换

令 \({t}_{1}=\ln {{x}_{1}}\)，\({t}_{2}=\ln {{x}_{2}}\)

由\(\ln {{x}_{1}}=a{{x}_{1}}\)， \(\ln {{x}_{2}}=a{{x}_{2}}\) 得 \({t}_{1}=a{e}^{{t}_{1}}\) ，\({t}_{2}=a{e}^{{t}_{2}}\)，\(\therefore \frac {{t}_{1}} {{t}_{2}}={e}^{{t}_{2}-{t}_{1}}\)

要证\({x}_{1}{x}_{2}>{e}^{2}⇔\ln {\left ( {{x}_{1}{x}_{2}} \right )>2⇔\ln {{x}_{1}}}+\ln {{x}_{2}}>2⇔{t}_{1}+{t}_{2}>2\)

令\(k=\frac {{t}_{2}} {{t}_{1}}>1\)，则 \({t}_{2}-{t}_{1}=\ln {k}\)

\(\therefore {t}_{1}=\frac {\ln {k}} {k-1}\)，\({t}_{2}=\frac {k\ln {k}} {k-1}\)，则\({t}_{1}+{t}_{2}=\frac {\left ( {k+1} \right )\ln {k}} {k-1}\)

只需证：\(\ln {k>\frac {2\left ( {k-1} \right )} {k+1}}\left ( {k>1} \right )\)同理法三

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