已知函数f(x)=x的平方+（2/X）+alnX(X>0),f(x)导函数是f'(x).对任意两个不等的正数X1，X2，证明：

{[f(X1)+f(X2)]/2}-f[(X1+X2)/2]
=(x1^2+2/x1+alnx1+x2^2+2/x2+alnx2)/2-[(x1+x2)^2/4+2/(x1+x2)/2+aln(x1+x2)/2]
=[(x1^2+x2^2）/2-(x1+x2)^2/4)]+[（2/x1+2/x2)/2-2/(x1+x2)/2）]+[（alnx1+alnx2）/2-aln(x1+x2)/2]

[(x1^2+x2^2）/2-(x1+x2)^2/4)]=(x1-x2)^2/4≥0
（2/x1+2/x2)/2-2/(x1+x2)/2）=(x1-x2)^2/[x1x2(x1+x2)]≥0
（alnx1+alnx2）/2-aln(x1+x2)/2=aln[√(x1x2)/(x1+x2)/2]
当x1，x2>0
显然(x1+x2)/2≥2√(x1x2)/2=√(x1x2)
所以ln[√(x1x2)/(x1+x2)/2]<=ln1=0
当a<=0时，aln[√(x1x2)/(x1+x2)/2]≥0
所以
{[f(X1)+f(X2)]/2}-f[(X1+X2)/2]≥0
则{[f(X1)+f(X2)]/2}>f[(X1+X2)/2]

|f'(x1)-f'(x2)|=2x1-2/x1^2+a/x1-(2x2-2/x2^2+a/x2)|
=|2(x1-x2)-2(1/x1^2-1/x2^2)+a*(1/x1-1/x2)|
=|2+2(x1+x2)/x1^2x2^2-a/x1x2||x1-x2|
而
|2+2(x1+x2)/x1^2x2^2-a/x1x2|
=|2(x1^2x2^2+2(x1+x2)-ax1x2))/x1^2x2^2|

x1^2x2^2+2(x1+x2)-ax1x2≥x1^2x2^2+4√(x1x2)-ax1x2
当0<x1x2<1时，
a≤4时。
4√(x1x2)>ax1x2
则x1^2x2^2+2(x1+x2)-ax1x2≥x1^2x2^2+4√