已知函数f(x)=x∧2+aln(x+1)证明:a=1时对任意的x1,x2∈【1,+∞),且x1≠x2都有【f（x1）-

【f（x1）-f(x2)】÷(x1-x2)={(x1^2-x2^2)+[ln(x1+1)-ln(x2+1)]}/(x1-x2)
=(x1-x2)*(x1+x2)/(x1-x2)+[ln(x1+1)-ln(x2+1)]/(x1-x2)
=(x1+x2)+[ln(x1+1)-ln(x2+1)]/(x1-x2)
(x1+x2)≥2
但不能证明[ln(x1+1)-ln(x2+1)]/(x1-x2)≥0.5
当x1>>x2,[ln(x1+1)-ln(x2+1)]/(x1-x2)-->0
但能证明[ln(x1+1)-ln(x2+1)]/(x1-x2)>0,因为当x>1,ln(1+x)递增.
【f（x1）-f(x2)】÷(x1-x2) >2