If $f(x)$ is smooth, or $f\in C^1$, assume $\displaystyle h(t)=\int_a^t f(s)\mathrm{d}s$, and $\displaystyle g(t)=\int_a^{t}\frac{1}{f(s)}\mathrm{d}s$,we just can focus on \begin{equation} \frac{h(t)}{\exp(2Lg(t)-1)} \end{equation}
because we know that there is a $\xi\in(c,d)$ s.t. \begin{equation} \frac{h(d)-h(c)}{g(d)-g(c)}=\frac{h'(\xi)}{g'(\xi)}=f^2(\xi) \end{equation}
thus we just prove $\frac{h(t)}{\exp(2Lg(t)-1)}\le\frac{\min_{[a,b]}f^2(\xi)}{2L}$
To find the minimum. Let's compute the derivative and we also can see the LHS function at $t=a$ has a limit as $f^2(a)/2L$.
Assume the LHS term is $\gamma(t)$, then \begin{equation} \gamma'(t)=\frac{h'(\exp(2Lg)-1)-2Lg'\exp(2Lg)h}{(\exp(2Lg)-1)^2} \end{equation}
It is easy to compute the limit at $t=a$, we find that $\gamma'(a)=\frac{f(a)f'(a)}{2L}-\frac{1}{2}f(a)\le0$. And we shall see that if we set \begin{eqnarray} p(t)&=&f\cdot(h'(\exp(2Lg)-1)-2Lg'\exp(2Lg)h)\\ &=&f^2(\exp(2Lg)-1)-2L\exp(2Lg)h \end{eqnarray}
We can see that $p(a)=0$, since $h(a)=0$ and $g(a)=0$. However, \begin{eqnarray} p'(t)&=&2ff'(\exp(2Lg)-1)-4L^2\exp(2Lg)h/f\\ &\le&\frac{2L}{f}(f^2(\exp(2Lg)-1)-2L\exp(2Lg)h)\\ &=&\frac{2L}{f}p(t) \end{eqnarray}
Thus we shall know that $p(t)\le 0$, since $\{(\exp(-2Lg(t))p\}'\le0$,and $p(a)=0$.
So we shall get that $\gamma'(t)\le 0$.
Thus $\gamma$ is decreasing in $t$, $\gamma(t)\le\gamma(a)$.
On the other hand. Take $\displaystyle\int_{t}^b f(s)\mathrm{d}s=\phi(t)$,$\displaystyle\int_{t}^b\frac{1}{f(s)}\mathrm{d}s=\psi(t)$.
Then we also can see that \begin{equation} \beta(t)=\frac{\phi(t)}{\exp(2L\psi(t)-1)} \end{equation} which has $\displaystyle\beta(b)=\frac{f^2(b)}{2L}.$
with the same process(PAY ATTENTION TO THE SIGN), $\beta'(b)=\frac{f(b)f'(b)}{2L}+\frac{f(b)}{2}\ge 0.$
And also we can obtain that $q(t)=2L \exp(2L \psi)\phi-f^2(\exp(2L\psi)-1)\ge 0$
which means $\beta(t)$ is increasing in $t$, which is $\beta(t)\le\beta(b)$.
Since $\beta(a)=\gamma(b)$, thus $\gamma(b)\le \min(f^2(a),f^2(b))/2L$.
AND the choice of $b$ is arbitrary, we know that for any $t$, we have $\gamma(t)\le f^2(t)/2L$
Consider the $\min_{[a,b]}f^2(t)$ is reached at $t=\upsilon$, then $\gamma(\upsilon)\le f^2(\upsilon)/2L$, since $\gamma(t)$ is decreasing in $t$. Thus $\gamma(b)\le f^2(\upsilon)/2L$. $\Box$