Prove that $\int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \geq 4$, under the given conditions on $f(x)$

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,
\[\text{max}_{[0,1]} f(x)\leq \frac{ab}{a+b}.\]
Furthermore,
\[ \int_0^1|f''(x)|d x \geq \left|\int_0^1f''(x) dx \right|\]
\[=|f'(1)-f'(0)|\]
\[=a+b. \]
Since $\min\limits_{\mathbb{R^+}}\frac{(1+t)^2}{t}=4$, we have 
\[ \int_0^1\left|\frac{f''(x)}{f(x)}\right|d x\]
\[\ge\frac{1}{\max\limits_{[0,1]}f(x)}\int_0^1|f''(x)|\,\mathrm{d}x\]
\[\ge\frac{(a+b)^2}{ab}\]
\[=\frac{(1+b/a)^2}{b/a}\geq 4.\]