已知a>0,a不等于1，m>n>0,比较A=a^m+1/a^m与B=a^n+1/a^n的大小

A-B=a^m+1/a^m-a^n-1/a^n =a^m-a^n+(1/a^m-1/a^n) =a^m-a^n+(a^n-a^m)/(a^m*a^n) =(a^m-a^n)(1-1/a^mn) =a^n * (a^(m-n)-1) * (1-1/a^mn) 变成上式之后可知在a>0,a不等于1的情况下A-B>0 故知A>B