Integral Inequality $|f''(x)/f(x)|$ Let $f$ be a $C^2$ function in $[0,1]$ such that $f(0)=f(1)=0$ and $f(x)\neq 0\,\forall x\in(0,1).$ Prove that
$$\int_0^1 \left|\frac{f{''}(x)}{f(x)}\right|dx\ge4$$
 A: Just some considerations. Without loss of generality, we can assume that $f(x)$ is positive on $(0,1)$. Let $\operatorname{graph}(g)$ be the convex hull of $\operatorname{graph}(f)$. We have $g(x)\in C^2([0,1])$ and
$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|g''(x)|}{g(x)}dx,$$
so we can also assume that $f$ is a concave function over $[0,1]$, with $f'(0)=\alpha>0,f'(1)=-\beta<0$ and a unique maximum in $x_0\in(0,1)$, for which $f'(x_0)=0$ and $f(x_0)=1$. Let now $\operatorname{graph}(h)$ be the envelope of the tangent lines to $\operatorname{graph}(f)$ for  $x\in\left\{0,x_0,1\right\}$. For any $\epsilon>0$ there is a function $u\in C^2([0,1])$ such that $|u-h|<\epsilon$ and:
$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|u''(x)|}{u(x)}dx=\alpha+\beta-O(\epsilon),$$
but $\alpha+\beta$ must be greater than $4$, since $\max u$ is below the $y$-coordinate of the intersection of the tangent lines in $x=0$ and $x=1$, so
$$ 1 = \max u \leq \frac{\alpha\beta}{\alpha+\beta} \leq_{AM-GM} \frac{\alpha+\beta}{4}.$$ 
A: Proving the Lower Bound
Without loss of generality, assume that $f(x)\gt0$ for $x\in(0,1)$.
Suppose that $f(x_0)=y_0=\max\limits_{x\in[0,1]}f(x)$. Then, $f'(x_0)=0$. 
By the Mean Value Theorem, for some $x_1\in(0,x_0)$, $f'(x_1)=\frac{y_0}{x_0}$. Therefore,
$$
\int_0^{x_0}|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{x_0}
$$
Furthermore, for some $x_2\in(x_0,1)$, $f'(x_2)=-\frac{y_0}{1-x_0}$. Therefore,
$$
\int_{x_0}^1|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{1-x_0}
$$
Since $f(x)\le y_0$,
$$
\begin{align}
\int_0^1\left|\,\frac{f''(x)}{f(x)}\,\right|\,\mathrm{d}x
&\ge\frac{\frac{y_0}{x_0}+\frac{y_0}{1-x_0}}{y_0}\\
&=\frac1{x_0}+\frac1{1-x_0}\\
&=\frac1{\frac14-\left(x_0-\frac12\right)^2}\\[3pt]
&\ge4
\end{align}
$$

The Lower Bound is Sharp
Let
$$
f_a(x)=\sin^{-1}\left(\frac{\sin(\pi x)}{1+a^2\sin^2(\pi x)}\right)
$$

then
$$
\lim_{a\to0}\int_0^1\left|\,\frac{f_a''(x)}{f_a(x)}\,\right|\,\mathrm{d}x=4
$$
since $f_a''(x)$ is tends to $0$ except near $x=\frac12$, and $\int_0^1f_a''(x)\,\mathrm{d}x$ tends to $-2\pi$, whereas $f_a\!\left(\frac12\right)$ tends to $\frac\pi2$.
