Prove that : $|f(b)-f(a)|\geqslant (b-a) \sqrt{f'(a) f'(b)}$ with $(a,b) \in \mathbb{R}^{2}$ 
Let $(a,b) \in \mathbb{R}^{2}$ such that $a<b$ and $f\in C^2([a,b],\mathbb{R})$ such that   $f'\neq 0$ and $f''/f'$ is decreasing. 
Prove that :
$$|f(b)-f(a)|\geqslant (b-a) \sqrt{f'(a) f'(b)}$$

My attempt:
By the MVT theorem, we have to prove : 
$$
|f'(c)| \geq \sqrt{f'(a)f'(b)}, \quad c\in (a,b)
$$
Without loss of generality we can assume that $f'$ is stricly positive.
Thus, by the fact that $\ln(f')$ is concave we get 
$$
\frac{\ln(f'(c))-\ln(f'(a))}{c-a}\geqslant\frac{\ln(f'(b)-\ln(f'(c))}{b-c}
$$
So,
$$
\frac{\ln \frac{f'(c)}{f'(a)}}{c-a}\geqslant\frac{\ln \frac{f'(b)}{f'(c)}}{b-c}
$$
Notice that I haven't use  $\frac{f''(a)}{f'(a)}\geq \frac{f''(b)}{f'(b)}$,
So I think what I am doing it's not enough.
Any ideas?
Thank you in advance for your help,

Abbreviate $f'=g$. Please note that you can not use the MVT, as
$$|g(c)|\geq \sqrt{g(a)g(b)},\quad \forall c\in (a,b) $$
is not possible if $g(a)\neq g(b)$. Instead we note that (wlog $g>0$)
$$|f(b)-f(a)|=\int_a^b g(x)dx=(b-a)\int_0^1g((1-s)a+bs)ds. $$
As $g$ is log concave, we have that
$$g(sb+(1-s)a)\geq g(b)^sg(a)^{1-s}.$$
This gives us
$$|f(b)-f(a)|\geq (b-a)\int_0^1g(b)^sg(a)^{1-s}ds =(b-a)\frac{g(b)-g(a)}{ln(g(b))-ln(g(a))}.$$
The inequality
$$ \frac{g(b)-g(a)}{ln(g(b))-ln(g(a))} \geq \sqrt{g(a)g(b)} $$
is not too difficult to proof, I will leave it to you. (Please ask if you need help).
