已知a+1大于0,比较a+1分之a与a+2分之a+1的大小

a+1>0,比较a/(a+1)与(a+1)/(a+2)的大小因为a+1>0,所以a+2>1[a/(a+1)]-[(a+1)/(a+2)]=[a(a+2)-(a+1)^2]/(a+1)(a+2)=(a^2+2a-a^2-2a-1)/(a+1)(a+2)=-1/(a+1)(a+2)(a+1)>0,(a+2)>1所以1/(a+1)(a+2)>0,-1/(a+1)(a+2)