证明：∵b＞a＞0,∴a+1＞1,则ln（a+1）＞0,b+1＞0,a+b+1＞0.即ln（a+1）／b+1＞0,而（a-b）（a+b+1）＜0.∴ln(a+1)／b+1(a-b)(a+b+1).
已知函数f(x)=-x^2+ln(1+2x),设b>a>0,证明：ln(a+1)/b+1>(a-b)(a+b+1)
证明：∵b＞a＞0,∴a+1＞1,则ln（a+1）＞0,b+1＞0,a+b+1＞0.即ln（a+1）／b+1＞0,而（a-b）（a+b+1）＜0.∴ln(a+1)／b+1(a-b)(a+b+1).