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,

{[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]1时,
则x1^2x2^2+2(x1+x2)-ax1x2≥x1^2x2^2+4√(x1x2)-ax1x2
=(x1x2-a/2)^2+4√(x1x2)-a^2/4
x1x2>1,a≤4时,4√(x1x2)-a^2/4>0
所以,(x1x2-2)^2+4√(x1x2)-4>=0
则x1^2x2^2+4√(x1x2)-ax1x2>=0
则x1^2x2^2+2(x1+x2)-ax1x2≥0
即2(x1^2x2^2+2(x1+x2)-ax1x2))>=x1^2x2^2
综合得：|f'(x1)-f'(x2)|>|x1-x2|

我给你回答第一问：
f'(x).=2x-1/x^2+a/x
f"(x)=2+2/x^3-a/x^2
=2+1/x^2*(2/x-a)
当a小于等于零时，f"(x)恒大于等于零；所以f(x)为凹函数，所以得证！