设函数f(x)=arctanx,g(x)=x,x＞0 f(0)=0,g(0)=0 f'(x)=1/(1+x²)＞0,g'(x)=1＞0 f'(x)-g'(x)=1/(1+x²)-1=-x²/(1+x²)≤0 即f'(x)≤g'(x) 因为[0,+∞)上f(x)与g(x)单调递增且f'(x)≤g'(x) 所以x>arctan(x)