g(x)=lnx-2(x-1)/(x+1) ,x小于1,gx小于0;x大于1 ,gx大于0 令hx=xf(x)=ax-xlnx-1,0小于x小于e^(a-1)=t时 hx递增,x大于t时 hx递减,所以g(x1/t)小于0 即lnx1小于lnt+2(x1-t)/(x1+t) 所以ax1-1=x1lnx1小于x1lnt+2x1(x1-t)/(x1+t) 即(a-lnt-2)x1^2+(at-1-tlnt+2t)x1-t小于0 即x1^2-(3e^(a-1)-1)x1+e^(a-1)大于0,同理 x2^2-(3e^(a-1)-1)x2+e^(a-1)小于0,所以(x1+x2)(x1-x2)大于(3^(e-1))(x1-x2) 所以x1+x2小于3e^(a-1)