Suppose that $f(x)$ has twice order continuous derivative over $[a,b]$, $\forall x \in [a,b]:f''(x)<0$ and $f(a)=f(b)=0.$ Prove $$\displaystyle \int_a^b \left|\frac{f''(x)}{f(x)}\right|{\rm d}x>\frac{4}{b-a}.$$
Proof
Since $f''(x)<0$, then $f(x)$ is concave over $[a,b]$. Hence, $\forall x \in (a,b):f(x)>0$. Moreover, since $f(x)$ is continuous over $[a,b]$, it can reach its maximum value at some point $x=c \in (a,b)$,namely $\max\limits_{x \in [a,b]}f(x)=f(c)>0$ and $f'(c)=0$. Therefore $$\int_a^b \left|\frac{f''(x)}{f(x)}\right|{\rm d}x=-\int_a^b \frac{f''(x)}{f(x)}{\rm d}x> -\int_a^b \frac{f''(x)}{f(c)}{\rm d}x=\frac{f'(a)-f'(b)}{f(c)}.\tag{0}$$ By Taylor's formula, we obtain $$f(c)=f'(a)(c-a)+\frac{f''(\xi_1)}{2}(c-a)^2<f'(a)(c-a)\tag{1}$$ $$f(c)=f'(b)(c-b)+\frac{f''(\xi_2)}{2}(c-b)^2<f'(b)(c-b)\tag{2}$$ From $(1)$, we obtain $$\frac{f'(a)}{f(c)}>\frac{1}{c-a}.\tag{3}$$ From $(2)$, we obtain $$-\frac{f'(b)}{f(c)}>\frac{1}{b-c}.\tag{4}$$ By $(3)+(4)$ $$\frac{f'(a)-f'(b)}{f(c)}>\frac{1}{c-a}+\frac{1}{b-c}.\tag{5}$$ By HM-AM inequality $$\frac{2}{\dfrac{1}{c-a}+\dfrac{1}{b-c}}\leq \frac{(c-a)+(b-c)}{2}=\frac{b-a}{2}.\tag{6}$$ which implies $$\frac{1}{c-a}+\frac{1}{b-c}\geq \frac{4}{b-a}.\tag{7}$$ Combining $(0)$ and $(7)$, it follows that $$\int_a^b \left|\frac{f''(x)}{f(x)}\right|{\rm d}x>\frac{4}{b-a},$$ which is what we want to prove.
The integral equals $\infty$ in all cases.
Proof: From concavity we see $f'(a)>0.$ Since concave functions stay below their tangent lines, we have $f(x) \le f'(a)(x-a)$ for $x\in [a,b].$ We are also given that $|f''| \ge m$ on $[a,b]$ for some $m>0.$ Thus
$$\int_a^b \left |\frac{f''(x)}{f(x)}\right |\, dx \ge \int_a^b \frac{m}{f'(a)(x-a)}\, dx=\infty.$$