设有\( \xi_i \in (0,1) \)，对f(x)在区间\([0,x_1]\),\([x_1,x_2]\),\([x_2,x_3]\),\( \cdots \),\([x_n,1]\),上分别使用拉格朗日中值定理可得
\[ f'(\xi_1) = \frac{f(x_1)-f(0)}{x_1-0} = \frac{f(x_1)}{x_1} \Rightarrow \exists \xi_1 \in (0,x_1) , \frac{1}{f'(\xi_1)} = \frac{x_1}{f(x_1)}\]
\[ f'(\xi_2) = \frac{f(x_2)-f(x_1)}{x_2-x_1} \Rightarrow \exists \xi_2 \in (x_1,x_2) , \frac{1}{f'(\xi_2)} = \frac{x_2-x_1}{f(x_2) – f(x_1)}\]
\[ \vdots\]
\[ f'(\xi_n) = \frac{f(1)-f(x_{n-1})}{1-x_{n-1}} \Rightarrow \exists \xi_n \in (x_{n-1},1) , \frac{1}{f'(\xi_n)} = \frac{1-x_{n-1}}{1- f(x_{n-1})}\]
要证\[ \frac{1}{f'(\xi_1)} + \frac{1}{f'(\xi_2)} + \cdots +\frac{1}{f'(\xi_n)} = n \]
只需\[ f(x_1) = \frac{1}{n},f(x_2) = \frac{2}{n}, \cdots, f(x_{n-1}) = \frac{n-1}{n} \]
又\(f(0)=0\),\(f(1)=1\)，根据连续函数的介值定理，存在\( x_k \in (0,1) \),\( k \in[1,n-1]\)，使\(f(x_k) = \frac{k}{n}\)成立，
故得证。