Inequality for integral s Suppose that $ f_1,f_2 : \mathbb{R}\rightarrow [0, \infty )$ are integrable functions and the ratio $f_1/f_2$ is a decreasing function. How do you prove that  $$g(x)=\frac{
\int_0^{x}f_1dt}
{ \int_0^x f_2 dt}$$ is a decreasing function?
 A: $$g(x)-g(x+\varepsilon) = \frac{\int_{0}^{x}f_1\,dt\int_{0}^{x+\epsilon}f_2\,dt-\int_{0}^{x}f_2\,dt\int_{0}^{x+\varepsilon}f_1\,dt}{\int_{0}^{x}f_2\,dt \int_{0}^{x+\varepsilon}f_2\,dt}$$
hence it is enough to prove that
$$ \int_{0}^{x}\int_{x}^{x+\varepsilon}f_1(u)\,f_2(v)-f_2(u)\,f_1(v)\,dv\,du > 0$$
or that:
$$ \int_{0}^{x}\int_{x}^{x+\varepsilon}\left(\frac{f_1}{f_2}(u)-\frac{f_1}{f_2}(v)\right)\,dv\,du > 0 $$
that is trivial since over the integration range $v>u$.
There is also an equivalent formulation for the ratio of the partial sums of two series with positive terms: I remember it is a crucial lemma in some inequalities about elliptic integrals.
A: Let $y>x$. We have 
$$g(x)-g(y)=\frac{\int_0^xf_1(t)dt}{\int_0^xf_2(t)dt}-\frac{\int_0^yf_1(u)du}{\int_0^yf_2(u)du} $$
To show that $g(x)-g(y)\geq 0$, we have only to show that
$$A(x,y)=(\int_0^xf_1(t)dt)(\int_0^yf_2(u)du)-(\int_0^xf_2(t)dt)(\int_0^y f_1(u)du) \geq 0$$
It is easy to see that
$$A(x,y)= (\int_0^xf_1(t)dt)(\int_x^yf_2(u)du)-(\int_0^xf_2(t)dt)(\int_x^y f_1(u)du)$$
We have
$$A(x,y)=\int\int_D(f_1(t)f_2(u)-f_2(t)f_1(u))dudt=\int\int_Dh(t,u)dudt$$
Where $D=\{(t,u); 0\leq t\leq x$ and $x\leq u\leq y$. As for $(t,u)\in D$, we have $u\geq t$, it is easy to see that $h(u,t)\geq 0$, and we are done.  
