non-trivial result about integrals due Gromov Gromov proved that if $$
f,g:\left[ {a,b} \right] \to R
$$
are integrable functions, such that the function $$
t \to \frac{{f\left( t \right)}}
{{g\left( t \right)}}
$$
is also integrable, and decreasing.  Then the function $$
r \to \frac{{\int\limits_a^r {f\left( t \right)dt} }}
{{\int\limits_a^r {g\left( t \right)dt} }}
$$
is decreasing.
I could not proved, and I could not find a proof )=
 A: Thanks to Mariano Suárez-Alvarez for pointing out a bad assumption I made in my previous attempt
For all $u\le v$, in $[a,b]$ we have
$$
\frac{f(u)}{g(u)}\ge\frac{f(v)}{g(v)}
$$
Assuming that $g$ is either non-negative or non-positive on all of [a,b], we get
$$
f(u)g(v)\ge f(v)g(u)
$$
Let $r\le s$. Then, integrating in $u$ from $a$ to $r$ and then in $v$ from $r$ to $s$, we get
$$
\int_a^rf(u)\mathrm{d}u\;\int_r^sg(v)\mathrm{d}v\ge\int_a^rg(u)\mathrm{d}u\;\int_r^sf(v)\mathrm{d}v
$$
Then we have
$$
\begin{align}
&\frac{\int_a^rf(t)\mathrm{d}t}{\int_a^rg(t)\mathrm{d}t}-\frac{\int_a^sf(t)\mathrm{d}t}{\int_a^sg(t)\mathrm{d}t}\\
&=\frac{\int_a^rf(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t-\int_a^rg(t)\mathrm{d}t\;\int_a^sf(t)\mathrm{d}t}{\int_a^rg(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t}\\
&=\frac{\int_a^rf(t)\mathrm{d}t\;(\int_a^rg(t)\mathrm{d}t+\int_r^sg(t)\mathrm{d}t)-\int_a^rg(t)\mathrm{d}t\;(\int_a^rf(t)\mathrm{d}t+\int_r^sf(t)\mathrm{d}t)}{\int_a^rg(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t}\\
&=\frac{\int_a^rf(t)\mathrm{d}t\;\int_r^sg(t)\mathrm{d}t-\int_a^rg(t)\mathrm{d}t\;\int_r^sf(t)\mathrm{d}t}{\int_a^rg(t)\mathrm{d}t\;\int_a^sg(t)\mathrm{d}t}\\
&\ge0
\end{align}
$$
Update: The requirement that $g$ stay either non-negative or non-positive is reasonable since the result is false for $f(t)=1-t$ and $g(t)=1-t^2$ on $[0,\frac{3}{2}]$. Here is the graph of $\frac{\int_0^x(1-t)\;\mathrm{d}t}{\int_0^x(1-t^2)\;\mathrm{d}t}$:

A: The geometric interpretation of the result is fairly clear if you draw the picture of a particle with velocity vector $(f(t), g(t))$ that at time $t=a$ is at $(0,0) \quad$ (assume $g(t) > 0$ so that the particle moves to the right at all times).  Decreasing $f(t)/g(t)$ means the path of the particle is convex, curving downward.  This implies the second property if the particle goes through $0$; the slope of the velocity vector when $t>a$ is always less than the slope of the line from the particle to $0$ so that continued motion forces the latter to decrease.
A: Extra assumption 1: $g$ is non-negative or non-negative. (Thanks robjohn)
Extra assumption 2: $f$ and $g$ are absolute continuous (e.g. they are strictly increasing/decreasing). (Thanks Mariano Suárez-Alvarez and t.b.)
Fix $r$. Since $f/g$ is decreasing we have 
$$\frac{f(x)}{g(x)}\ge\frac{f(r)}{g(r)}$$
for all $a\le x\le r$. Hence (by Extra assumption 1)
$$f(x)g(r)\ge f(r)g(x)$$
next we integrate this with respect to $x$ over $[a,r]$ which leads to
$$g(r)\int_a^rf(x)dx\ge f(r)\int_a^rg(x)dx$$
(recall $r$ was fixed). But then (by Extra assumption 2) 
$$\left(\frac{\int_a^rf(x)dx}{\int_a^rg(x)dx}\right)'= \frac{f(r)\int_a^rg(x)dx-g(r)\int_a^rf(x)dx}{\left(\int_a^rg(x)dx\right)^2}\le0.$$
