Show that $f''(\xi)≥8$ 
Let $f:[0,1]\rightarrow\mathbb{R}$ be a $\mathcal{C}^{2}([0,1])$ function such that $f(0)=f(1)=1$ and such that the minimum of $f$ is negative. Prove that there must exist a point $\xi\in [0,1]$ such that $f''(\xi)≥8$

Maybe I can use the fact that there exists maximum and minimum in $[0,1]$ so there are two critical points. But I don't know how to proceed. Can someone give me any advice? Thanks before!
 A: I can get to $4$, by mean value theorem.
Proof:
Let $x_{\min}$ denote the minimum point. By mean value theorem there is $c_1,c_2$ such that
$$f'(c_1)=\frac{f(x_{\min})-f(0)}{x_{\min}},f'(c_2)=\frac{f(1)-f(x_{\min})}{1-x_{\min}}.$$
And then, by mean value theorem there is $c_3$:
$$f''(c_3)=\frac{f'(c_2)-f'(c_1)}{c_2-c_1}$$
$$=\frac{\frac{f(1)-f(x_{\min})}{1-x_{\min}}-\frac{f(x_{\min})-f(0)}{x_{\min}}}{c_2-c_1}$$
$$>{\frac{1}{1-x_{\min}}+\frac{1}{x_{\min}}}$$
$$=\frac{1}{x_{\min}(1-x_{\min})}\geq 4$$
You can likely get a tighter bound following the comments.
EDIT: The correct way...
By Taylor's theorem, there is $c$ such that
$$f(x)=f(x_{min})+\frac{f''(c)(x-x_{min})^2}{2}.$$
Then
$$f''(c)=2\frac{f(x)-f(x_{min})}{(x-x_{min})^2}$$
Now develop this at $x=0,x=1$ and sum those, then there is $c_1$, $c_2$
$$\frac{f''(c_1)+f''(c_2)}{2}=\frac{1-f(x_{min})}{x_{min}^2}+\frac{1-f(x_{min})}{(1-x_{min})^2}>\frac{1}{x_{min}^2}+\frac{1}{(1-x_{min})^2}$$
$$=\frac{1-2x_{min}+2x_{min}^2}{(1-x_{min})^2x_{min}^2}\geq8$$
which can only hold if $f''(c_1)$ or $f''(c_2)$ is greater than $8$.
A: Assume $f'(x)<8$ for all $x\in(0,1)$.
At ist minimum point $x_\min$, we have
$$f(x_\min)<0,\qquad f'(x_\min)=0. $$
Then for $x>x_\min$,
$$ f'(x)=f'(x)-f'(x_\min)=\int_{x_\min}^xf''(t)\,\mathrm dt<8(x-x_\min)$$
whereas for $x<x_\min$,
$$ f'(x)=f'(x)-f'(x_\min)=-\int_x^{x_\min}f''(t)\,\mathrm dt>8(x-x_\min).$$
By integrating again,
$$\tag1 1<f(1)-f(x_\min)=\int_{x_\min}^1f'(x)\,\mathrm dx<8\int_{x_\min}^1(x-x_\min)\,\mathrm dx=4(1-x_\min)^2$$
and
$$\tag2 1<f(0)-f(x_\min)=-\int_0^{x_\min}f'(x)\,\mathrm dx<8\int_0^{x_\min}(x_\min-x)\,\mathrm dx=4x_\min^2.$$
Multiply $(1)$ and $(2)$ and use AM-GM inequality:
$$ 1<16\left(\sqrt{(1-x_\min)(x_\min)}\right)^4<16\left(\frac{(1-x_\min)+x_\min}2\right)^4=1,$$
contradiction.
