.已知a>1,m>n>0,比较A=a^m+1/a^m和B=a^n+1/a^n的大小

A=a^m+1/a^m=(a^2m+1)/a^mB=a^n+1/a^n=(a^2n+1)/a^nA-B=(a^2m+1)/a^m-(a^2n+1)/a^n=(a^2m*a^n+a^n-a^2n*a^m-a^m)/a^m*a^n分母大于0看分子a^2m*a^n+a^n-a^2n*a^m-a^m=a^m*a^n(a^m-a^n)-(a^m-a^n)=(a^m-a^n)(a^m*a^n-1)a>1,m>0,所以a^m>1同理a^n>1所以a^m*a^n>1a^m*a^n-1>0a>1,所以a^x是增函数m>n所以a^m-a^n>0所以分子也大于0所以A>B