Prove that $\int_0^1 \left| \frac{f^{''}(x)}{f(x)} \right| dx \geq4$.

Here's another answer, which avoids reducing to the case in which $f$ is concave. It is plainly enough to prove that $$ \sup_{x \in [0,1]}{|f(x)|} \leq \frac{1}{4}I,\quad \text{where} \quad I := \int_0^1|f''(x)|\,dx. \tag{1} $$ First of all, since $f(0) = f(1) = 0$ and $f$ is nonzero in $(0,1)$, we know that the supremum on the left-hand side is attained at some $c \in (0,1)$, and moreover, that $f'(c) = 0$. By using Taylor's theorem with remainder (which is really just repeated integration by parts) to expand $f$ around the point $c$, we have $$ f(x) = f(c) + f'(c)(x-c) + \int_c^x (x - t)f''(t)\,dt = f(c) + \int_0^c (x - t)f''(t)\,dt, $$ for any $x \in [0,1]$. Successively taking $x = 0$ and $x = 1$ gives $$ f(c) = -\int_0^c tf''(t)\,dt = -\int_c^1 (1-t)f''(t)\,dt, $$ because $f(0) = f(1) = 0$. This means that $$ |f(c)| \leq c\int_0^c |f''(t)|\,dt \quad \text{and} \quad |f(c)| \leq (1-c) \int_c^1 |f''(t)|\,dt. \tag{2} $$ Since $$ \int_0^c|f''(t)|\,dt + \int_c^1|f''(t)|\,dt = I = (1-c)I + cI, $$ we must have either $\int_0^c|f''(t)|\,dt \leq (1-c) I$ or $\int_c^1|f''(t)|\,dt \leq c I$. Either way $(2)$ shows that $$ |f(c)| \leq c(1-c)I = \frac{1}{4}I - \left(\frac{1}{2} - c\right)^2I \leq \frac{1}{4}I, $$ and $(1)$ is therefore proved.

Without loss of generality, assume that $f(x)\ge0$ on $[0,1]$. Furthermore, we can assume that $f''(x)\le0$. If not, we can replace $f$ by $g$ where the graph of $g$ is the convex hull of the graph of $f$. Note that where $g(x)\not=f(x)$, $g''(x)=0$, therefore, $\int_0^1\left|\frac{g''(x)}{g(x)}\right|\,\mathrm{d}x\le\int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x$.

Suppose that $f'(0)=a$ and $f'(1)=-b$. Since $f$ is concave, $f(x)\le ax$ and $f(x)\le b(1-x)$. Therefore, $$ \max_{[0,1]}f(x)\le\frac{ab}{a+b}\tag{1} $$ Furthermore, $$ \begin{align} \int_0^1|f''(x)|\,\mathrm{d}x &\ge\left|\int_0^1f''(x)\,\mathrm{d}x\right|\\[6pt] &=|f'(1)-f'(0)|\\[6pt] &=a+b\tag{2} \end{align} $$ Therefore, since $\min\limits_{\mathbb{R^+}}\frac{(1+t)^2}{t}=4$, $$ \begin{align} \int_0^1\left|\frac{f''(x)}{f(x)}\right|\,\mathrm{d}x &\ge\frac{1}{\max\limits_{[0,1]}f(x)}\int_0^1|f''(x)|\,\mathrm{d}x\\ &\ge\frac{(a+b)^2}{ab}\\ &=\frac{(1+b/a)^2}{b/a}\\ &\ge4\tag{3} \end{align} $$

There is an elementary proof for this inequation, by mean-value theorem.

CASE I: $f(x)\geq 0 $ on $[0,1]$

$f(0)=f(1)=0$ There is a point $x_0\in (0,1)$ s.t $\max_{[0,1]}f(x)=f(x_0)$

By mean-value theorem ,there are $\lambda_1 \in (0,x_0),\lambda_2 \in (x_0,1)$

s.t $\begin{align} f(x_0)-f(0)=\int_0^{x_0} f'(x) dx = (x_0 - 0)f'(\lambda_1)\Rightarrow f'(\lambda_1)=\frac{f(x_0)}{x_0}\end{align}$

$\begin{align} f(1)-f(x_0)=\int_{x_0}^{1} f'(x) dx = (1-x_0)f'(\lambda_2)\Rightarrow f'(\lambda_2)=\frac{-f(x_0)}{1-x_0}\end{align}$

$\begin{align} \int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \ge \int_{\lambda_1}^{\lambda_2} \left| \frac{f''(x)}{f(x)} \right| dx \ge \int_{\lambda_1}^{\lambda_2} \frac{|f''(x)|}{f(x_0)} dx \\= \frac {1}{f(x_0)} \int_{\lambda_1}^{\lambda_2} |f''(x)| \ge \frac {1}{f(x_0)} |\int_{\lambda_1}^{\lambda_2} f''(x) |\end{align}$

And $\begin{align} \frac {1}{f(x_0)}| \int_{\lambda_1}^{\lambda_2} f''(x)| =\frac {1}{f(x_0)}|f'(\lambda_2)-f'(\lambda_1)| =\frac {1}{f(x_0)}|\frac{-f(x_0)}{1-x_0}-\frac{f(x_0)}{x_0}|=\frac {1}{(1-x_0)(x_0)}\ge 4\end{align}$

CASE II: $f(x)$ may be nagetive on $[0,1]$.

Since $\forall x\in [0,1],|f(x)|\le M$ for some M,let $g(x) = f(x) + 2M$.

Then $\forall x\in [0,1],|g(x)|\ge M \ge |f(x)|$ and $g''(x)=f''(x),g(x)\ge 0$ on $[0,1]$

$\begin{align} \int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx =\int_0^1 \left| \frac{g''(x)}{f(x)} \right| dx\ge \int_0^1 \left| \frac{g''(x)}{g(x)} \right| dx \ge 4\end{align} $ (By CASE I)