Prove $\int_a^b f(x)dx \leq \frac{e^{2L\beta}-1}{2L\alpha}\int_c^d f(x)dx$ Assume $f(x)>0$ defined in $[a,b]$, and for a certain $L>0$, $f(x)$ satisfies the Lipschitz condition $|f(x_1)-f(x_2)|\leq L|x_1-x_2|$.
Assume that for $a\leq c\leq d\leq b$,$$\int_c^d \frac{1}{f(x)}dx=\alpha,\int_a^b\frac{1}{f(x)}dx=\beta$$Try to prove$$\int_a^b f(x)dx \leq \frac{e^{2L\beta}-1}{2L\alpha}\int_c^d f(x)dx$$
 A: I got something which is rather close to your result but couldn't get rid of an additional term. I'm hoping someone will find the development useful in order to give a complete answer. The question reminded me somehow of the proof of Gronwall's inequality and my answer is based on that.
Let $h(t)=\int_a^t f(s)ds$ taking the derivative with respect to $t$ yields
$h'(t) = f(t)$. Rewriting $f(t)=f^2(t)/f(t)$, we can find an upper bound for $f^2(t)$ indeed, we have:
$$f^2(t) = \int_a^t (f^2(s))'ds + f^2(a) = 2\int_a^t f(s)f'(s)ds +f^2(a)$$
but then, the Lipschitz continuity implies that $|f'(s)|\le L$ which yields
$$f^2(t) \le 2L \int_a^t f(s)ds + f^2(a)\quad \Longrightarrow \quad h'(t) \le \left(2Lh(t) +f^2(a)\right)\left({1\over f(t)}\right) $$
we can then introduce the additional positive term $\left(\int_c^df(t)dt\right)/\alpha$ with:
$$ h'(t) \le \left(2L h(t)+f^2(a) + {\int_c^d f(s)ds\over \alpha}\right)\left({1\over f(t)}\right) $$
Remark: the additional term comes from the $f^2(a)$. We could get rid of it if $\left(\int_c^d f(s)ds\right)/\alpha > f^2(a)$ which I wasn't able to prove.
To simplify notations, let $a=f^2(a) + {\int_c^d f(s)ds\over \alpha},$ $b=2L$ and $g(t)=1/f(t)$ then. Then the previous inequality reads
$$h'(t) \le g(t)(a+bh(t))$$
which can be rewritten as follows (here is where it starts to look like Gronwall's inequality):
$$ {(a+bh(t))'\over a+bh(t)} \le bg(t) $$
the left-hand side is the logarithmic derivative of $a+bh(t)$, integrating both sides from $a$ to $t$ yields
$$ a+bh(t) \le a\exp\left(b\int_a^t g(t) \right) $$
plugging the values of $a$,$b$ and using $\int_a^b g(t) = \beta$, we finally get:
$$\int_a^b f(s)ds = h(b) \le \color{green}{\left[{\exp(2L\beta)-1\over 2L\alpha}\right]\int_c^d f(s)ds} + \color{red}{{f^2(a)\over 2L}\left(\exp(2L\beta)-1\right)} $$
A: 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 $\gamma(b)\le f^2(\upsilon)/2L$.
$\Box$
