A function on [a,b] that is second differentiable and f'(a)=f'(b)=0 Let $f:[a,b]\rightarrow\mathbb{R}$ be secondly differentiable and $f'(a)=f'(b)=0$. Then there exits a point $c\in [a,b]$ such that $$|f''(c)|\geq\frac{4}{(b-a)^2}|f(b)-f(a)|.$$
I tried to prove it by contradition, however, I didn't make it. I don't know how comes to the term on the right hand side of the inequality.
 A: I need to clean this up... but here is the idea.
There is a point in the interval where f(x) crosses the straight line between (a,f(a)) and (b,f(b)).  
i.e. if f(b) > f(a), then f(x) is below the line near a and above the line near b.  Therefore it crosses the line in between.
There are points on either side of this intersection where f'(x) is parallel to said line. call them p and q.
This now cuts our interval into 3 sub intervals.
Either (a,p) or (q,b) is less than (or equal to) 1/4 (a,b)
or, (p,q) is less than (or equal to) 1/2 (a,b).
Suppose $(a,p) \le 1/4 (a,b):$
then $\exists c \in (a,p)$ such that $f(c) = \frac{f'(p)}{(p-a)} \le \frac{4f(b)-f(a)}{(b-a)}$ 
Similar for $(q,b) \le 1/4 (a,b).$
Otherwise. $(q-p) < 1/2 (b-a), f'(p) = f'(q) = \frac {f(b)-f(a)}{b-a}$
Which is a very similar situation as where we began this problem.
I am going to leave it for now.  Let me know if you get this across the finish line.
A: To keep it simple, I'll work on $[0,1].$
Lemma: Suppose $g$ is differentiable on $[0,1]$ and $|g'|\le 1$ on $[0,1].$ If $g(0) = g(1) = 0,$ then $|\int_0^1 g| < 1/4.$
Proof: On the interval $[0,1/2],$ $|g(x)| \le x$ by the mean value theorem. Similarly, $|g(x)|\le 1-x$ on $[1/2,1].$ So the graph of $g$ is confined to the diamond defined by the points $(0,0), (1/2,1/2), (1,0), (1/2,-1/2).$ Now $\int_0^1 g = \int_0^1 \max (g,0) + \int_0^1 \min (g,0).$ The first integral is no more than the area of the upper triangle in the diamond, which is $1/4.$ In fact it cannot equal $1/4.$ Otherwise the graph of $g$ coincides with the upper two sides of the diamond. That violates differentiability at $1/2.$ Thus $0\le \int_0^1 \max (g,0) < 1/4.$ Similarly $-1/4 <\int_0^1 \min (g,0) \le 0.$ The sum of these integrals therefore lies in $(-1/4,1/4),$ proving the lemma.
Now suppose $\sup_{[0,1]}|f''| = 1$  and $f'(0)=f'(1) = 0.$ By the lemma we get $|\int_0^1 f'| < 1/4.$ But $\int_0^1 f' = f(1)-f(0).$ Thus
$$\tag 1 4\frac{|f(1) - f(0)|}{(1-0)^2} < 1.$$
Since $\sup |f''| = 1,$ we know by Darboux that there exists $c$ such that $|f''(c)|$ is at least the the fraction in $(1).$ In fact, we've shown there exists $c$ such that $|f''(c)|$ is greater than that.
For the general result you can scale the above, both in the range and domain.
