Showing a bound of the second derivative using remainder in integral form Let $a,b \in (0,\infty)$ Let $f$ be twice continuously differentiable , $f(0)=f'(0)=f'(a)=0, f(a)=b.$
Show that there exists $c \in (0,a)$, such that $$\vert f''(c) \vert \geq \frac{4b}{a^2}$$
I have to use the Taylor theorem with integral remainder to prove this.
 A: I have a result differing by a factor $2$ from yours.
The Taylor theorem with integral remainder is
$$
f(x) = f(x_0) + f'(x_0)(x-x_0) + \int_{x_0}^x f''(t)(x-t)dt
$$
Taking $x_0=0$ and $x=a$ we get
$$
f(a) = f(0) + f'(0) a + \int_0^a f''(t)(a-t)dt
$$
that becomes
$$
b = \int_0^a f''(t)(a-t)dt
$$
Given that $f''$ is continuous, then $|f''|$ has a maximum $|f''(c)|=M$ in $c\in[0,a]$, such that
$$
b = |b| = \left|\int_0^a f''(t)(a-t)dt\right| \leq \int_0^a \left|f''(t)(a-t)\right|dt \leq M\int_0^a(a-t)dt 
  = M\frac{1}{2}a^2
$$
Then
$$
|f''(c)|=M\geq\frac{2b}{a^2}
$$

Edit
As suggested by @Bruno B exchanging the role of $0$ and $a$ we get
$$
f(0) = f(a) + f'(a)(0-a)+\int_a^0f''(t)(0-t)dt
$$
i.e.
$$
b = - \int_0^a f''(t)tdt
$$
and adding the two relation
$$
2b = \int_0^a f''(t)(a-2t)dt \leq M \int_0^a|a-2t|dt = \frac{1}{2}Ma^2
$$
and so
$$
M \geq \frac{4b}{a^2}
$$
A: Since the original question without stating the necessity of using Taylor got deleted, here is my previous answer again.
First note that using the rescaling
$$
g(x)=\frac{f(ax)}{b}
$$
we can assume wlog that $a=b=1$.
To get to a contradiction assume that we have $f\in C^2(\mathbb{R})$, $f(0)=f'(0)=f'(1)=0,\,f(1)=1$ such that $|f''|<4$, i.e. $-4<f''<4$, everywhere.
Then on the one hand we have
\begin{align*}
f\bigg(\frac{1}{2}\bigg)
&=\int_0^\frac{1}{2}f'(t)\,\mathrm{d}t+f(0)\\
&=\int_0^\frac{1}{2}\int_0^tf''(s)\,\mathrm{d}s\,+f'(0)\,\mathrm{d}t\\
&=\int_0^\frac{1}{2}\int_0^tf''(s)\,\mathrm{d}s\,\,\mathrm{d}t\\
&<\int_0^\frac{1}{2}\int_0^t4\,\mathrm{d}s\,\,\mathrm{d}t=\frac{1}{2}
\end{align*}
while one the other hand we have
\begin{align*}
f\bigg(\frac{1}{2}\bigg)
&=f(1)-\int_\frac{1}{2}^1f'(t)\,\mathrm{d}t\\
&=1-\int_\frac{1}{2}^1f'(1)-\int_t^1f''(s)\,\mathrm{d}s\,\mathrm{d}t\\
&=1+\int_\frac{1}{2}^1\int_t^1f''(s)\,\mathrm{d}s\,\,\mathrm{d}t\\
&>1-\int_\frac{1}{2}^1\int_t^14\,\mathrm{d}s\,\,\mathrm{d}t=\frac{1}{2},
\end{align*}
a contradiction.
A: Vincenzo edited their answer before finished writing this, but oh well, it's not a race! Instead I'll write what I had in mind, which was slightly different but not by much.
We can apply the Taylor formula with integral remainder twice given the conditions that $f$ satifies: once on $f$ and once on $f'$.
We get:
$$\begin{align*} b = f(a) &= f(0) + f'(0)a + \int_{0}^{a} f''(t)(a-t)\mathrm{d}t = \int_{0}^{a} f''(t)(a-t)\mathrm{d}t\\ 0 = f'(a) &= f'(0) + \int_{0}^{a} f''(t) \mathrm{d}t = \int_{0}^{a} f''(t) \mathrm{d}t\end{align*}$$
We can therefore obtain that, for all $\lambda \in \mathbb{R}$ :
$$b = b + \lambda \cdot 0 = \int_{0}^{a} f''(t)(a-t)\mathrm{d}t + \lambda \int_{0}^{a} f''(t)\mathrm{d}t = \int_{0}^{a} f''(t)(a + \lambda -t)\mathrm{d}t$$
Just like Vincenzo did earlier, we can then consider $c \in [0,a]$ such that $|f''(c)| = M = \max_{[0,a]} |f''|$:
$$ b = |b| = \left|\int_{0}^{a} f''(t)(a + \lambda -t)\mathrm{d}t\right| \leq \int_{0}^{a} \left|f''(t)(a + \lambda -t)\right|\mathrm{d}t \leq M\int_{0}^{a} |a + \lambda - t|\mathrm{d}t$$
Right off the bat, we see that we only gain information here if we look at $\lambda \in [-a,0]$.
However, after a few simplifications, we find that, for that given $\lambda$ range:
$$\int_{0}^{a} |a + \lambda - t|\mathrm{d}t = \frac{a^2}{2} + \lambda a + \lambda^2$$
Seen as a polynomial function $g$ in $\lambda$, we know its minimum is in $\frac{-a}{2 \cdot 1} = -\frac{a}{2}$, which provides the minimum $g\left(-\frac{a}{2}\right) = \frac{a^2}{4}$, which lets us conclude the same way as Vincenzo, except that this consideration about all possible $\lambda$ indicates that this should be the best bound with the Taylor formula with integral remainder (plus we could have sped up the process by "guessing" $\lambda = -\frac{a}{2}$ from the get-go).
