已知函数\(g(x)={\ln}x-\dfrac{1}{2}mx-1\).
\((1)\) 讨论\(g(x)\)的单调性;
\((2)\) 若函数\(f(x)=xg(x)\)在\((0,+\infty)\)上存在两个极值点\(x_1,x_2\),且\(x_1<x_2\),证明\(:{\ln}x_1+{\ln}x_2>2\).
解析:
\((1)\) 对函数\(g(x)\)求导可得\[
g'(x)=\dfrac{1}{x}-\dfrac{1}{2}m,x>0.\]
情形一 若\(m\leqslant 0\),则\(g(x)\)单调递增.
情形二 若\(m>0\),则\(g(x)\)在\(\left(0,\dfrac{2}{m}\right)\)上单调递增,在\(\left[\dfrac{2}{m},+\infty\right)\)单调递减.
\((2)\) 由题\[
f(x)=x{\ln}x-\dfrac{1}{2}mx^2-x,x>0.\]对\(f(x)\)求导可得\[
f'(x)={\ln}x-mx,x>0.\]
从而\(x_1,x_2\)是\(f'(x)\)的两个变号零点,因此\[
m=\dfrac{{\ln}x_1}{x_1}=\dfrac{{\ln}x_2}{x_2}=\dfrac{{\ln}x_1+{\ln}x_2}{x_1+x_2}=\dfrac{{\ln}x_1-{\ln}x_2}{x_1-x_2}>\dfrac{2}{x_1+x_2}.\]
于是\({\ln}x_1+{\ln}x_2>2\)得证.
来源：https://www.cnblogs.com/Math521/p/11827392.html