Prove that $\int\limits_0^1 \bigg | \frac{f''(x)}{f(x)} \bigg| dx \ge \frac{4(M-1)}{M}$

Let $f:[0,1]\to \mathbb{R}$ be a $C^2$ class function such that $f(0)=f(1)=1$ and $f(x)>1,\forall x\in (0,1)$.
Prove that $$\int\limits_0^1 \bigg | \frac{f''(x)}{f(x)} \bigg| dx \ge \frac{4(M-1)}{M},$$ where $M=\max\limits_{x\in [0,1]}f(x).$
I can't solve this problem, but here are some of the things I have tried/observed:

• $0$ and $1$ are extreme points of $f$, but we cannot apply Fermat's theorem since they are the endpoints of $f$'s domain

• $$\int\limits_0^1 \bigg | \frac{f''(x)}{f(x)} \bigg| dx \ge \int\limits_0^1 \bigg |\frac{f''(x)}{f'(x)}\cdot \frac{f'(x)}{f(x)}\bigg | dx\ge \bigg|\int\limits_0^1 \frac{f''(x)}{f'(x)}\cdot \frac{f'(x)}{f(x)} dx \bigg|$$ Then, I tried to apply Integration by Parts, but to no avail.

• $\exists m\in (0,1)$ such that $f(m)=M$. From Lagrange's MVT on $[0,m]$ and $[m,1]$, \exists $a\in (0,m), b\in (m,1)$ such that $$M=f(m)=mf'(a) \space \text{and} \space M-1=f(m)-1=(m-1)f'(b).$$
  Now, we have that $$\int\limits_0^1 \bigg | \frac{f''(x)}{f(x)} \bigg| dx \ge \frac{1}{M}\int\limits_0^1 |f''(x) |dx \ge \frac{1}{M} \int\limits_a^b |f''(x)|dx \ge \frac{1}{M} \bigg|\int\limits_a^b f''(x) dx \bigg |=$$$$=\frac{1}{M}(f'(b)-f'(a))$$ If I substitute $f'(b)$ and $f'(a)$ in terms of $m$ and $M$, I cannot prove the required inequality.
  EDIT: It should be $M-1$=mf'(a) $. With this, the problem is solved if we proceed the way I did.

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