Define $f$ has second continuous derivative over $[0,1]$. And $f(0)=f(1)=0$, And $f(x)\neq0 \,when \ x\in(0,1)$, prove $$\int_{0}^{1} |\frac{f{''}(x)}{f(x)}|\,dx\ge 4,$$
I can't use $f(0)=0 \ and\ f(1)=0$ completely at the same time whether using the Lagrange Mean Value Theorem or the Fundamental Theorem of Calculus. So I can not solve it. How to prove that? Thank you.
WLOG, assume that $f(x) > 0$ on $(0, 1)$.
$f$ achieves its global maximum $M > 0$ on $[0, 1]$, at some point $x_0\in (0, 1)$.
By MVT, we have $f(x_0) - f(0) = f'(t_1)x_0$ and $f(1) - f(x_0) = f'(t_2)(1-x_0)$ for some $t_1 \in (0, x_0)$ and $t_2\in (x_0, 1)$ which results in $f'(t_1) = \frac{M}{x_0}$ and $f'(t_2) = \frac{-M}{1-x_0}$.
Then, we have \begin{align} \int_0^1 \frac{|f''(x)|}{|f(x)|}\mathrm{d}x &\ge \int_{t_1}^{t_2} \frac{|f''(x)|}{|f(x)|}\mathrm{d}x\\ &\ge \frac{1}{M}\int_{t_1}^{t_2} |f''(x)|\mathrm{d}x\\ &\ge \frac{1}{M}\left|\int_{t_1}^{t_2} f''(x)\mathrm{d}x \right|\\ &= \frac{1}{M}|f'(t_2)-f'(t_1) |\\ &= \frac{1}{M}\left|\frac{-M}{1-x_0}-\frac{M}{x_0}\right|\\ &= \frac{1}{x_0(1-x_0)}\\ &\ge \frac{1}{(x_0 + 1-x_0)^2/4}\\ &= 4. \end{align}
Remarks: $4$ is the best constant. To see this, consider the following example. For $0 < \epsilon < \frac{1}{2}$, let
\begin{align} f(x) = \left\{\begin{array}{ll} x & 0\le x \le \tfrac{1}{2} - \epsilon \\[5pt] ax^4+bx^3+cx^2+dx+e & \tfrac{1}{2}-\epsilon < x < \tfrac{1}{2} + \epsilon \\[5pt] 1-x & \tfrac{1}{2} +\epsilon \le x \le 1 \end{array} \right. \end{align} where $a = \frac{1}{8\epsilon^3}, b = - \frac{1}{4\epsilon^3}, c = \frac{3 - 12\epsilon^2}{16\epsilon^3}, d = \frac{12\epsilon^2-1}{16\epsilon^3}$, and $e = \frac{1-24\epsilon^2+64\epsilon^3-48\epsilon^4}{128\epsilon^3}$.
For each fixed $\epsilon \in (0, 1/2)$, $f(x)$ satisfies the hypotheses. As $\epsilon \to 0$, $\int_0^1 \frac{|f''(x)|}{|f(x)|}\mathrm{d}x \to 4$.