已知函数 $f(x)=x \ln x$, 若 $0<x_{1}<x_{2}$, 则下列结论正确的是 ( )
设 $g(x)=rac{f(x)}{x}=\ln x$, 函数单调递增, 则 $g\left(x_{2}
ight)>g\left(x_{1}
ight)$即 $rac{fleft(x_{2}
ight)}{x_{2}}>rac{f\left(x_{1}
ight)}{x_{1}} 	herefore x_{1} f\left(x_{2}
ight)>x_{2} f\left(x_{1}
ight), A$ 正确;设 $h(x)=f(x)+x 	herefore h^{\prime}(x)=\ln x+2$ 不是恒大于零, $B$ 错误;$f(x)=x \ln x 	herefore f^{\prime}(x)=\ln x+1$ 不是恒小于零, $C$ 错误;$\ln x>-1$ 故 $f^{\prime}(x)=\ln x+1>0$, 函数单调递增故 $\left(x_{2}-x_{1}
ight)\left(f\left(x_{2}
ight)-f\left(x_{1}
ight)
ight)=x_{1} f\left(x_{1}
ight)+x_{2} f\left(x_{2}
ight)-x_{2} f\left(x_{1}
ight)-x_{1} f\left(x_{2}
ight)>0$,即 $x_{1} f\left(x_{1}
ight)+x_{2} f\left(x_{2}
ight)>x_{2} f\left(x_{1}
ight)+x_{1} f\left(x_{2}
ight)$,$rac{f\left(x_{2}
ight)}{x_{2}}=\ln x_{2}>rac{f\left(x_{1}
ight)}{x_{1}}=\ln x_{1} 	herefore x_{1} f\left(x_{2}
ight)>x_{2} f\left(x_{1}
ight)$, 即 $x_{1} f\left(x_{1}
ight)+x_{2} f\left(x_{2}
ight)>2 x_{2} f\left(x_{1}
ight), D$ 正确.