证明f(x)=√(x^2+1)-x在定义域内是减函数
f(x1)-f(x2)>0
√(x1^2+1)-x1-√(x2^2+1)+x2>0
√(x1^2+1)-√(x2^2+1)>x1-x2
[√(x1^2+1)-√(x2^2+1)]^2>(x1-x2)^2
x1^2+1+x2^2+1-2√[(x1^2+1)(x2^2+1)]>x1^2-2x1x2+x2^2
2-2√[(x1^2+1)(x2^2+1)]>-2x1x2
1+x1x2>√[(x1^2+1)(x2^2+1)]
(1+x1x2)^2>[(x1^2+1)(x2^2+1)]
1+x1^2*x2^2+2x1x2>x1^2*x2^2+x1^2+x2^2+1
1+2x1x2>x1^2+x2^2+1
x1^2-2x1x2+x2^2
(x1-x2)^2>0
所以 减函数