用拉格朗日中值定理证明 e的X方>=1+X

令 f(x) = e^x,
x = 0 时， 显然有 e^x = 1+x.
任给 x > 0，
f(x) - f(0) = f'(t)(x-0) --- 0 = e^t * x > x
=> e^x = f(x) > x + f(0) = 1 + x;
任给 x f(0) - f(x) = f'(t)(0-x) --- x = e^t * (-x) => e^x = f(x) > x + f(0) = 1 + x;

2的X方=(1+1)的X方,广义二项式展开.........>1+x,即2的X方>1+x，因为e>2,所以e的X方>=1+X

令f(x)=e^x-x-1 f(x)满足拉格朗日中值定理.
f(0)=0
f(x)-f(0)=f'(ξ)x
f'(x)=e^x-1 当x>=0时,f'(x)>=0
f(x)-f(0)>=0 问题得证;
当x0
f(x)-f(0)>=0 问题得证.

e^x - 1 - x = e^x - e^0 - x = (e^ξ)x - x = (e^ξ - e^0)x = (e^ξ')ξx
其中，ξ在0和x之间，与x同号，ξ'在0和ξ之间，所以(e^ξ')ξx ≥ 0