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Prove $\int_0^1 \left| \frac{f^{''}(x)}{f(x)} \right| dx \geq4$

Suppose $f\in C^2[0, 1]$, $f(0) = f(1) = 0$. For any $x\in(0,1)$, $f(x)\neq 0$.

Please show $\int_0^1|\frac{f^{\prime\prime}(x)}{f(x)}|dx\geq 4$.

Here is what I got.

It is clear the sign of $f(x)$ keeps fixed. WLOG, assume $f(x)>0$ for any $x\in(0,1)$. Say $f$ attains its maximum $A$ at $c\in(0,1)$. We have $\frac{A}{c} = \frac{f(c)-f(0)}{c-0}=f^{\prime}(a)$, $a\in(0,c)$; and $\frac{-A}{1-c}=\frac{f(1)-f(c)}{1-c}=f^{\prime}(b)$, $b\in(c,1)$.

Then we just need to show $\int_0^1\frac{|f^{\prime\prime}(x)|}{f(x)}dx\geq\int_a^b\frac{|f^{\prime\prime}(x)|}{f(x)}dx\geq\frac{1}{A}\int_a^b|f^{\prime\prime}(x)|dx\geq4$.

And we also know there is some $d\in(a,b)$, such that $|f^{\prime\prime}(d)|=|\frac{f^{\prime}(b)-f^{\prime}(a)}{b-a}|=\frac{\frac{A}{c(1-c)}}{b-a}$

Then I do not know how to proceed. Can someone help me to get rid of this difficulty?

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    I think I got it. $\int_0^1\frac{|f^{\prime\prime}(x)|}{f(x)}dx\geq\frac{1}{A}\int_a^b|f^{\prime\prime}(x)|dx\geq\frac{1}{A}|\int_a^bf^{\prime\prime}(x)dx|=\frac{1}{A}|f^{\prime}(b)-f^{\prime}(a)|=\frac{1}{c(1-c)}\geq 4$ because $c\in(0,1)$.2012-03-14

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