If nonnegative $f: [0,1] \rightarrow \mathbb{R}$ has a continuous $f''$, then $\int_0^1 \Big| \frac{f''(x)}{f(x)} \Big| \,dx >4$ Assume that $f: [0,1] \rightarrow \mathbb{R}$ has a continuous $f''$ and $f$ is positive on the interval $(0,1)$ and $0$ at the endpoints.  
I want to prove that $$\int_0^1 \Big| \frac{f''(x)}{f(x)} \Big| \,dx >4. $$
I could not make any progress. Since the unit interval is compact, we know that the continuity of $f, f'$ and $f''$ are uniform and they attain their maximum and minimum values. Moreover, I tried to apply the Mean Value Theorem but I failed. 
 A: Since, $f$ is positive on the interval $(0,1)$ it suffices to prove that,
$\displaystyle \int_0^1 |f''(t)|\,dt \ge 4\max\limits_{x \in [0,1]} |f(x)|$
Since, $f(0) = f(1) = 0$ using integration by parts we have:
$\displaystyle f(x) = -(1-x)\int_0^x tf''(t)\,dt - x\int_x^1 (1-t)f''(t)\,dt$ for all $x \in [0,1]$.
Denoting $\displaystyle \phi(t) := \begin{cases} (x-1)t & \textrm{ for } t \in [0,x] \\ x(t-1) & \textrm{ for } t \in [x,1]\end{cases}$
We can write the above identity as, $\displaystyle f(x) = \int_0^1 \phi(t)f''(t)\,dt$
Thus, $\displaystyle |f(x)| \le \max\limits_{t \in [0,1]} |\phi(t)|\int_0^1 |f''(t)|\,dt = x(1-x)\int_0^1 |f''(t)|\,dt \le \frac{1}{4}\int_0^1 |f''(t)|\,dt$
for all $x \in [0,1]$.
Hence, $\displaystyle \int_0^1 |f''(t)|\,dt \ge 4\max\limits_{x \in [0,1]} |f(x)|$ as desired.
Integration by parts:
$\displaystyle f(x) - f(0) = \int_0^x f'(t)\,dt = xf'(x) - \int_0^x tf''(t)\,dt\tag{1}$ 
$\displaystyle \begin{align} f(1) - f(x) = \int_x^1 f'(t)\,dt &= f'(1) - xf'(x) - \int_x^1 tf''(t)\,dt \\ &= (1-x)f'(x) + \int_x^1 (1-t)f''(t)\,dt \tag{2} \end{align}$
Multiplying $(1)$ with $(1-x)$ and $(2)$ with $x$, and subtracting, we get:
$$\displaystyle f(x) = -(1-x)\int_0^x tf''(t)\,dt - x\int_x^1 (1-t)f''(t)\,dt$$
