1. ## An Intergral Inequlity

if $\displaystyle f(x)\in C([0,1])$ and $\displaystyle 0 \leq f(x) < 1$

then
$\displaystyle \int_0^1\frac{f(x)}{1-f(x)}\geq\frac{\int_0^1f(x)}{1-\int_0^1f(x)}$

thanks very much

2. Originally Posted by Xingyuan

if $\displaystyle f(x)\in C([0,1])$ and $\displaystyle 0 \leq f(x) < 1$

then
$\displaystyle \int_0^1\frac{f(x)}{1-f(x)}\geq\frac{\int_0^1f(x)}{1-\int_0^1f(x)}$

thanks very much
Since $\displaystyle 0 \leq f(x) < 1$ we may apply the geometric series

$\displaystyle \frac{{f\left( x \right)}} {{1 - f\left( x \right)}} = f\left( x \right) + f^2 \left( x \right) + ...$

Now I'll show that: $\displaystyle \int_0^1 {f^m \left( x \right)dx} \geqslant \left( {\int_0^1 {f\left( x \right)dx} } \right)^m$ valid $\displaystyle \forall m \in \mathbb{Z}^ +$

Consider the Riemannian sum: $\displaystyle \int_0^1 {f^m \left( x \right)dx} = \mathop {\lim }\limits_{n \to + \infty } \left( {\tfrac{1} {n}} \right) \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k} {n}} \right)}$

By the Power Mean inequality we have: $\displaystyle \sqrt[m]{{\left( {\tfrac{1} {n}} \right) \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k} {n}} \right)} }} \geqslant \left( {\tfrac{1} {n}} \right) \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k} {n}} \right)}$

Thus:$\displaystyle \tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k} {n}} \right)} \geqslant \left( {\tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k} {n}} \right)} } \right)^m$ and: $\displaystyle \mathop {\lim }\limits_{n \to + \infty } \tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {f^m \left( {\tfrac{k} {n}} \right)} \geqslant \left( {\mathop {\lim }\limits_{n \to + \infty } \tfrac{1} {n} \cdot \sum\limits_{k = 1}^n {f\left( {\tfrac{k} {n}} \right)} } \right)^m$

So we have: $\displaystyle \int_0^1 {f^m \left( x \right)dx} \geqslant \left( {\int_0^1 {f\left( x \right)dx} } \right)^m$

This may also be shown applying Hölder's Inequality

Assuming uniform convergence *

$\displaystyle \int_0^1 {\frac{{f\left( x \right)}} {{1 - f\left( x \right)}} \cdot dx} = \int_0^1 {f\left( x \right)} \cdot dx + \int_0^1 {f^2 \left( x \right) \cdot } dx + ...$

Applying the inequality we've just proved: $\displaystyle \int_0^1 {\frac{{f\left( x \right)}} {{1 - f\left( x \right)}} \cdot dx} \geqslant \int_0^1 {f\left( x \right) \cdot dx} + \left( {\int_0^1 {f\left( x \right) \cdot dx} } \right)^2 + ...$

Which yields: $\displaystyle \int_0^1 {\frac{{f\left( x \right)}} {{1 - f\left( x \right)}} \cdot dx} \geqslant \frac{{\int_0^1 {f\left( x \right) \cdot dx} }} {{1 - \int_0^1 {f\left( x \right) \cdot dx} }}$

This last step is correct since: $\displaystyle 0 \leqslant \int_0^1 {f\left( x \right) \cdot dx} < 1$

By the way, there's equality iff the function is constant.

3. Originally Posted by PaulRS
Assuming uniform convergence *

$\displaystyle \int_0^1 {\frac{{f\left( x \right)}} {{1 - f\left( x \right)}} \cdot dx} = \int_0^1 {f\left( x \right)} \cdot dx + \int_0^1 {f^2 \left( x \right) \cdot } dx + ...$

$\displaystyle \frac{{f\left( x \right) - f^{t + 1} \left( x \right)}} {{1 - f\left( x \right)}} = f\left( x \right) + f^2 \left( x \right) + ... + f^t \left( x \right)$

Thus: $\displaystyle \int_0^1 {\frac{{f\left( x \right)}} {{1 - f\left( x \right)}}dx} - \int_0^1 {\frac{{f^{t + 1} \left( x \right)}} {{1 - f\left( x \right)}}dx} = \int_0^1 {f\left( x \right)dx} + ... + \int_0^1 {f^t \left( x \right)dx}$

$\displaystyle 0 \leqslant f\left( x \right) \leqslant k < 1 \Rightarrow 0 \leqslant \frac{{f^{t + 1} \left( x \right)}} {{1 - f\left( x \right)}} \leqslant \frac{{k^{t + 1} }} {{1 - k}}$ ( $\displaystyle g\left( x \right) = \tfrac{x} {{1 - x}}$ is increasing in $\displaystyle \left( {0,1} \right)$ )

Integrating: $\displaystyle \Rightarrow 0 \leqslant \int_0^1 {\frac{{f^{t + 1} \left( x \right)}} {{1 - f\left( x \right)}}} dx \leqslant \frac{{k^{t + 1} }} {{1 - k}}$

And since $\displaystyle \frac{{k^{t + 1} }} {{1 - k}} \to 0$ we have that: $\displaystyle \int_0^1 {\frac{{f^{t + 1} \left( x \right)}} {{1 - f\left( x \right)}}dx} \to 0$ as $\displaystyle t \to + \infty$

4. excellent proof. very powerful.

5. Originally Posted by Xingyuan
excellent proof. very powerful.

If you want a powerful result apply Jensen's inequality to prove that:

$\displaystyle \int_0^1f(g(x))dx\geq{f\left(\int_0^1g(x)dx\right) }$ where $\displaystyle f$ (a continuous function) is convex on the interval $\displaystyle [a,b]$; $\displaystyle g:[0,1]\rightarrow{[a,b]}$

This is the more general result, but I still like the proof above.