<s>[INST] Integral inequality with two increasing functions Let $f,g:[0,1]\rightarrow\mathbb{R}^+$ be increasing functions such that $f\leq g$. Is there a constant $c>0$ (independent of $f,g$) for which there exists some $r\geq 0$ (possibly dependent on $f,g$) such that
$$\int_{x:f(x)\leq r\leq g(x)}g(x)dx+\int_{f(x)\geq r}f(x)dx\geq c\int_0^1g(x)dx ?$$
As an example, let $g(x)=x$ and $f(x)=x^2$. The integral $\int_0^1g(x)dx$ is $\frac{1}{2}$. On the left-hand side, for fixed $r$, the first integral is from $x=r$ to $x=\sqrt{r}$ and amounts to $\frac{1}{2}(r-r^2)$. The second integral is from $x=\sqrt{r}$ to $x=1$ and amounts to $\frac{1}{3}(1-r\sqrt{r})$.
Suppose $g(x)=x^t$ and $f(x)=x^s$ for $s\geq t\geq 1$. Then the right-hand side is $\frac{1}{t+1}$. The left-hand side is
$$\frac{r^{\frac{t+1}{s}}}{t+1}-\frac{r^{\frac{t+1}{t}}}{t+1}+\frac{1}{s+1}-\frac{r^{\frac{s+1}{s}}}{s+1}.$$ 
 [/INST] $$
\newcommand {\diff} {\mathrm d}
\newcommand{\spc}[1] {\quad #1 \quad}
$$
For convenience, rewrite the inequality as $\max\limits_r \frac A B \ge c$, where
$$A = \int\limits_{f(x) \leq r \leq g(x)} g(x)\; \diff x + \int\limits_{f(x) \geq r} f(x)\; \diff x, \quad B = \int\limits_0^1g(x)\; \diff x$$
For $f(x) = \frac 1 {1 + s(1 - x)}$ and $g(x) = \frac 1 {1 + t(1 - x)}$, $s > t > 1$:
\begin{align*}
\frac A B
 & \spc = \left( \int\limits_{1 + \frac 1 t \left( 1 - \frac 1 r \right)}^{1 + \frac 1 s \left( 1 - \frac 1 r \right)} \frac {\diff x} {1 + t(1 - x)} + \int\limits_{1 + \frac 1 s \left( 1 - \frac 1 r \right)}^1 \frac {\diff x} {1 + s(1 - x)} \right) \Bigg/ \int\limits_0^1 \frac {\diff x} {1 + t(1 - x)} \\\\
 & \spc = -\; \frac {\ln \Big( \frac t s + \left( 1 - \frac t s \right) r \Big) + \frac t s \ln r} {\ln (1 + t)} \qquad \bigg| \qquad r \in \left[ {{\frac 1 {1 + t}}, 1} \right] \\
\end{align*}
We can exclude values of $r$ below $g(0) = \frac 1 {1 + t}$, as these are not good cadidates for maximizing $\frac A B$, for mostly obvious reasons.  Thus, $\max\limits_r \frac A B$ occurs at one of $r = \frac 1 {1 + t}$, $r = 1$, or some value of $r$ such that $\frac \diff {\diff r} \frac A B = 0$.
\begin{align*}
\frac \diff {\diff r} \frac A B
 & \spc = -\; \frac 1 {\ln (t + 1)} \left( \frac {s - t} {t + (s - t) r} + \frac t {sr} \right) \spc = 0 \qquad \bigg| \qquad r = \frac {t^2} {t^2 - s^2} < 0 \\
\end{align*}
The solution for $\frac \diff {\diff r} \frac A B = 0$ is spurious, as it lies outside $\left[ {{\frac 1 {1 + t}}, 1} \right]$.  Thus, we need only consider $r = \frac 1 {1 + t}$ and $r = 1$ for $\max\limits_r \frac A B$:
\begin{align*}
\frac A B \bigg|_{r = 1} & \spc = 0 \\
\frac A B \bigg|_{r = \frac 1 {1 + t}}
 & \spc = -\; \frac {\ln \Big( \frac t s + \left( 1 - \frac t s \right) \frac 1 {1 + t} \Big) + \frac t s \ln \frac 1 {1 + t}} {\ln (1 + t)}
   \spc = \frac t s + 1 -\; \frac {\ln \left(\frac {t^2} s + 1 \right)} {\ln (t + 1)} \\
\end{align*}
Obviously, $\max\limits_r \frac A B = \frac A B \big|_{r = \frac 1 {1 + t}} = \frac t s + 1 -\; \frac {\ln \left(\frac {t^2} s + 1 \right)} {\ln (t + 1)}$.
If we put $k = \frac t s$, $k \in (0, 1)$,
\begin{align*}
\max \frac A B & \spc = 1 + k -\frac {\ln (k t + 1)} {\ln (t + 1)}
 \spc \to k - \frac {\ln k} {\ln t} \spc \to k \qquad \bigg| \qquad t \to \infty
\end{align*}
We can choose $k$ arbitrarily close to 0, and then, by the above, choose $t$ such that $\max\limits_r \frac A B$ will be arbitrarily close to $k$, for the corresponding pair of $f(x) = \frac 1 {1 + \frac t k (1 - x)}$ and $g(x) = \frac 1 {1 + t(1 - x)}$.  Hence, we can choose $f$ and $g$ such that $\max\limits_r \frac A B$ will be arbitrarily close to $0$.
</s>