题160912（14分）若函数$fleft( x ight)={{e}^{x}}-aln x-a$（$a>e$）有两个零点${{x}_{1}},{{x}_{2}}$，求证：${{x}_{1}}+{{x}_{2}}>2$．
证明：函数$y=fleft( x ight)$的零点，即方程$dfrac{1}{a}={{ ext{e}}^{-x}}left( ln x+1 ight)$的根，
设$gleft( x ight)={{ ext{e}}^{-x}}left( ln x+1 ight)$，因此${{x}_{1}},{{x}_{2}}$是方程$gleft( x ight)=dfrac{1}{a}$的两个实根．
$gleft( x ight)$的导数$g'left( x ight)={{e}^{-x}}left( dfrac{1}{x}-ln x-1 ight)$，
当$xin left( 0,1 ight)$时，$g'left( x ight)>0$，$gleft( x ight)$单增；当$xin left( 1,+infty ight)$上时，$g'left( x ight)<0$，$gleft( x ight)$单减，
$x=1$是$gleft( x ight)$的极大值点．
故方程$gleft( x ight)=dfrac{1}{a}$的两个实根一个比$1$大，一个比$1$小，不妨设${{x}_{1}}<1<{{x}_{2}}$．
构造函数$hleft( x ight)=-dfrac{1}{ ext{e}}{{left( x-1 ight)}^{2}}+dfrac{1}{ ext{e}}$，令$Fleft( x ight)=gleft( x ight)-hleft( x ight)$，
则${F}'left( x ight)={g}'left( x ight)-{h}'left( x ight)={{ ext{e}}^{-x}}left( dfrac{1}{x}-ln x-1 ight)+dfrac{2}{ ext{e}}left( x-1 ight)$，
于是${{F}'}'left( x ight)={{ ext{e}}^{-x}}left( -dfrac{1}{{{x}^{2}}}-dfrac{2}{x}+ln x+1 ight)+dfrac{2}{ ext{e}}$，
因为$y=-dfrac{1}{{{x}^{2}}}-dfrac{2}{x}+ln x+1$在$left( 0,+infty ight)$上单调递增，
所以当$xin left( 0,1 ight)$时，${{F}'}'left( x ight)<-2{{ ext{e}}^{-x}}+dfrac{2}{ ext{e}}<-2{{ ext{e}}^{-1}}+dfrac{2}{ ext{e}}=0.$
当$xin left( 1,+infty ight)$时，$F''left( x ight)>-2{{ ext{e}}^{-x}}+dfrac{2}{ ext{e}}>-2{{ ext{e}}^{-1}}+dfrac{2}{ ext{e}}=0,$
所以${F}'left( x ight)$在$left(0,1 ight)$上单调递减，在$left(1,+infty ight)$上单调递增，且${F}'left( 1 ight)=0$，
所以${F}'left( x ight)0$，$xin left( 0,+infty ight)$．
所以$Fleft(x ight)$单调递增，又$Fleft(1 ight)=0$，所以有$gleft(x ight)-hleft(x ight)egin{cases}>0,x>1,\=0,x=1,\<0,0<x<1.end{cases}$
设$y=dfrac{1}{a}$与二次函数$hleft(x ight)$交于$x_3,x_4$两点，不妨设$x_3<1<x_4$．因此，有
$dfrac{1}{a}=hleft( {{x}_{3}} ight)=gleft( {{x}_{1}} ight)<hleft( {{x}_{1}} ight)$，$dfrac{1}{a}=hleft( {{x}_{4}} ight)=gleft( {{x}_{2}} ight)>hleft( {{x}_{2}} ight)$，
如图所示：
所以$x_3<x_1<1<x_4<x_2,$
因为$x_3+x_4=2$，所以${{x}_{1}}+{{x}_{2}}>2$，得证．
注：本解析是通过构造二次函数$hleft(x ight)$，比较$gleft(x ight)$与$hleft(x ight)$在极值点两侧的大小关系，结合函数的单调性，得到$x_1,x_2,x_3,x_4$的大小关系，从而由$hleft(x ight)$与$y=dfrac{1}{a}$交点的横坐标之和为$2$，得到$x_1+x_2$与$2$的大小关系．难点是如何构造这样的二次函数，$gleft(x ight)$的极值点为$1$，则此二次函数$hleft(x ight)$需要满足$hleft( 1 ight)=gleft( 1 ight)$，${h}'left( 1 ight)={g}'left( 1 ight)$，${{h}'}'left( 1 ight)={{g}'}'left( 1 ight)$，
所以$hleft( x ight)=dfrac{{g}'left( 1 ight)}{2}{{left( x-1 ight)}^{2}}+gleft( 1 ight).$