How to prove $\frac{1}{2}f''(\xi) = \frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}$ Assume $f(x)$ is continuous in $[a,b]$, and $f''$ in $(a,b)$, prove that for every $c\in
(a,b)$, $\exists$ $\xi \in(a,b)$ such that     
$$\dfrac{1}{2}f''(\xi)=\dfrac{f(a)}{(a-b)(a-c)}+\dfrac{f(b)}{(b-a)(b-c)}+\dfrac{f(c)}{(c-a)(c-b)}.$$
I don't know how to use Cauchy Mean Theorem to prove it. 
 A: I assume that $f$ is continuous on $[a,b]$ and twice differentiable on $(a,b)$.  Let
$$
P(x):=f(x)-\left(f(a)\dfrac{(x-b)(x-c)}{(a-b)(a-c)}+f(b)\dfrac{(x-a)(x-c)}{(b-a)(b-c)}+f(c)\dfrac{(x-a)(x-b)}{(c-a)(c-b)}\right).
$$
Then, apply Rolle's Theorem successively to $P$ and $P'$.
[In more detail: $P$ is continuous on $[a,b]$, twice differentiable on $(a,b)$, and $P(a)=P(c)=P(b)=0$.  Therefore, by Rolle's Theorem, there exist $d$ in $(a,c)$ and $e$ in $(c,b)$ such that $P'(d)=P'(e)=0$.  Since $P'$ is differentiable on $(a,b)$, it is continuous on $[d,e]$, and it is differentiable on $(d,e)$.  So, we may apply Rolle's Theorem to $P'$ to find $\xi$ in $(d,e)$ with $P''(\xi)=0$.  This $\xi$ is then as desired.]
A: @David Moews
I have just come out the solution using Cauchy mean Theorem.                                                      $\dfrac{1}{2} f^{''}(\xi)=\dfrac{f(a)(c-b)+f(b)(a-c)+f(c)(b-a)}{(a-b)(b-c)(c-a)}$ 
so $\dfrac{1}{2} f^{''}(\xi)=\dfrac{f(a)(c-b)+f(b)(a-c)+f(c)(b-a)}{(a-b)(b-c)(c-a)}=\dfrac{f^{'}(\zeta_1)(a-b)+f(b)-f(a)}{(a-b)(2\zeta_1-a-b)}=\dfrac{f^{'}(\zeta_2)(a-b)+f(b)-f(a)}{(a-b)(2\zeta_2-a-b)}=\dfrac{(f^{'}(\zeta_1)-f^{'}(\zeta_2))(a-b)}{(a-b)(2\zeta_1-2\zeta_2)}$
where $a<\zeta_1<c<\zeta_2<b$ 
Then use Lagrange Theorem and that is                                                                  
