已知定义在(0,1)上的函数f(x)=(2^x)/(4^x+1)求证：函数f(x)在（0,1)上是单调递减

设0＜x1＜x2＜1
f(x2)－f(x1)＝2^(x2)/[4^(x2)＋1]－2^(x1)/[4^(x1)＋1]
＝[2^(x2)×4^(x1)＋2^(x2)－2^(x1)×4^(x2)－2^(x1)]/﹛[4^(x2)＋1]×[4^(x1)＋1]﹜
＝﹛[2^(x2)×4^(x1)－2^(x1)×4^(x2)]＋[2^(x2)－2^(x1)]﹜/﹛[4^(x2)＋1]×[4^(x1)＋1]﹜
＝﹛[1－2^(x1x2)[2^(x2)－2^(x1)]﹜/﹛[4^(x2)＋1]×[4^(x1)＋1]﹜
∵0＜x1＜x2＜1 ∴1＜2^(x1)＜2^(x2)＜2
∴2＜2^(x1x2)＝2^(x1)＋2^(x2)＜4 ∴1－2^(x1x2)＜0
∴f(x2)－f(x1)＝﹛[1－2^(x1x2)[2^(x2)－2^(x1)]﹜/﹛[4^(x2)＋1]×[4^(x1)＋1]﹜＜0
∴f(x2)＜f(x1)
∴函数f(x)在（0,1)上是单调递减