inequality of integral ratio $a(x),b(x),c(x),$ and $d(x)$ are positive function of $x$. $\frac{a(x)}{b(x)}$ and $\frac{c(x)}{d(x)}$ increases in $x$. Moreover, we have $\frac{a(x)}{b(x)}<\frac{c(x)}{d(x)}$ holds for all x $\in \Omega=[0,\bar{x}]$. Can we show the following inequality?
\begin{equation}
\frac{\int_\Omega a(x)\,\mathrm{d}x}{\int_\Omega b(x)\,\mathrm{d}x} <\frac{\int_\Omega c(x)\,\mathrm{d}x}{\int_\Omega d(x)\,\mathrm{d}x} 
\end{equation}
If not, what is the sufficient condition?
 A: There is no reason whatsoever (other than pure wishful thinking) for the inequality to hold true.
Hint: The discrete case is well known to be false. Look at Simpson's Paradox.
A: With not much care in picking the "best" counterexample, take, for instance,
$a(x)=x(1+x)$, $b(x)=x$, $c(x)=1$ and $d(x)=e^{-x}$. Then $a(x)/b(x)=1+x$ is increasing and $c(x)/d(x)=e^x$ is increasing as well. Additionally, $a/b\leq c/d$ since $1+x\leq e^x$. All your specified conditions are met.
Now compute
$$RHS = \frac{\int_0^a x+x^2 dx}{\int_0^a x dx}=\frac{\frac{1}{2}a^2+\frac{1}{3}a^3}{\frac{1}{2}a^2}=1+\frac{2}{3}a$$
and
$$LHS=\frac{\int_0^a 1 dx}{\int_0^a e^{-x}dx}=\frac{a}{1-e^{-a}}.$$
If we pick $a=1$, then numerical computations show that $RHS \approx 1.66$ and $LHS \approx 1.59$, which violates the inequality.
I do wonder if you required $a(x),b(x),c(x)$ and $d(x)$ to be increasing as well, maybe even bounded below, if your inequality will hold. In this case, the trick I used to get a counterexample would not work.
A: I know this is pretty late, but just came across a similar situation in research... Also, this answer will not be rigorous, coming from a physicist... But,

If $a(x), b(x), c(x), d(x)$ are functions over $x\in[l, u]$, satisfying:


*

*$b(x) > 0, d(x) > 0 ~~~\forall x$

*$\frac{a}{b} \leq \frac{c}{d} ~~~\forall x$

*$\frac{c}{d}$ is non-decreasing in $x$

*$\frac{d}{b}$ is non-decreasing in $x$
Then (assuming all four integrals are finite),
$$\frac{\int\limits_l^u~dx~a(x)}{\int\limits_l^u~dx~b(x)} \leq \frac{\int\limits_l^u~dx~c(x)}{\int\limits_l^u~dx~d(x)}$$

"Proof":
$$LHS = \frac{\int\limits_l^u~dx~b(x)~~a/b}{\int\limits_l^u~dx~b(x)}$$
$$\leq \frac{\int\limits_l^u~dx~b(x)~~c/d}{\int\limits_l^u~dx~b(x)}~~~~(\text{since } a/b \leq c/d)$$
$$\leq \frac{\int\limits_l^u~dx~b(x)~~d/b~~c/d}{\int\limits_l^u~dx~b(x)~~d/b}~~~~(\text{explained below})$$
$$=\frac{\int\limits_l^u~dx~c(x)}{\int\limits_l^u~dx~d(x)} = RHS$$
To see why the third line works, note that the second line represents the expectation value of $c(x)/d(x)$ over $x$ sampled according to a probability distribution function proportional to $b(x)$. This interpretation is allowed because $b(x) \geq 0$ and has a finite and positive integral.
The third line represents the weighted expectation value of the same quantity $c(x)/d(x)$ over $x$ sampled according to the same probability distribution function. The only difference is that now $c/d$ is weighted by $d/b$ (allowed because $d$ is also non-negative with a finite and positive integral).
Because c/d and d/b are both non-decreasing, this weighted average assigns higher weightage to higher values of c/d (sloppy phrasing). So the weighted average in line 3 is $\geq$ the unweighted average in line 2.

The result will continue to hold if condition (3) is replaced by
3*. $a/b$ is non-decreasing in $x$
A: A sufficient condition, which isn't very practical, is to say that ${a(x_1)}/{b(x_2)} < {c(x_3)}/{d(x_4)}$ for any $x_1,x_2,x_3,x_4 \in \Omega$. The result then follows from the first order Talyor expansion of the functions $A(x) = \int_0^x a(t) dt$, etc.
A little silly, I know, but it works!
