f(x)=[(x/m) - 1]^2+[(n/x)-1]^2的定义域[m,n],且1 <= m < n <= 2.(1)讨论函数f(x)的单调性；（２）证明：对任意的实数X1,X2∈[m,n],不等式｜f(X1)-f(X2)｜< 1恒成立．
热心网友
(1) 设x1,x2在定义域[m,n],且x1x2f(x1)-f(x2)=(x1-x2)/m*[(x1+x2)/m-2]-n(x1-x2)/x1x2*[n(x1+x2)/x1x2-2]=(x1-x2)[(x1+x2)/m^2-2/m-(x1+x2)n^2/x1^2x2^2+2n/x1x2]=(x1-x2)(1/m-n/x1x2)[(x1+x2)(1/m+n/x1x2)-2]x1-x20,1/m-n/x1x2=(x1x2-mn)/mx1x20所以f(x1)-f(x2)<0, f(x)递减