Some preliminary inequality

Consider a ratio $\int_A f(x,a)dx/\int_B f(x,a)dx$ where $A, B \subset [0,1]$ and $a \in R$. Suppose for any $x' \in A$ and $x \in B$, $f(x',a)/f(x,a) > f(x',b)/f(x,b).$ Then can we say that $\int_A f(x,a)dx/\int_B f(x,a)dx>\int_A f(x,b)dx/\int_B f(x,b)dx$? Thank you very much.

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What are your thoughts? What have you tried? You have posted 10 questions without accepting a single answer. Many don't seem to have much motivation-why do you care about this? –  Ross Millikan Mar 27 '11 at 23:41
What do you mean by "accepting an answer?" I am just a newbie and learning the rules. –  Thales Mar 27 '11 at 23:47
I have tried to understand by setting $A$ and $B$ are finite. Then I think I can show this inequality. So my thought is that it will hold. But I would like to know your thought. –  Thales Mar 27 '11 at 23:51
There is a gray checkmark near each answer. If you think an answer is the best one, click the one near it. It will turn green and give points to the answerer (and a few to you). –  Ross Millikan Mar 27 '11 at 23:56
Yes - they come out of my research. If necessary I can explain the motivation. –  Thales Mar 27 '11 at 23:57

Fix $y\in B$ therefore for any $x\in A$ $$f(x,a)>f(x,b)\frac{f(y,a)}{f(y,b)}$$ taking integral over $A$ implies $$\int_{A}f(x,a)dx>\int_{A}f(x,b)dx\left(\frac{f(y,a)}{f(y,b)}\right).$$ Thus$$f(y,b)\int_{A}f(x,a)dx>f(y,a)\int_{A}f(x,b)dx .$$ Now it is time to take integral over $B$, hence$$\int_{B}f(y,b)dy\int_{A}f(x,a)dx>\int_{B}f(y,a)dy\int_{A}f(x,b)dx ,$$ consequently $$\frac{\int_{A}f(x,a)dx}{\int_{B}f(y,a)dy}>\frac{\int_{A}f(x,b)dx}{\int_{B}f(y,b)dy}.$$