How to deal with the following integral inequality From Shengyin Recently, I am considering a question, as is well known, Cauchy's inequality is a famous and useful inequality. $$\left|\int_{a}^{b}f(x)dx\right|^2\leq|b-a|  \int_{a}^{b}f^2(x)dx.$$  My question is: can we obtain a inequality such that
$$\left|\int_{a}^{b}f(x)dx\right|^2\ge |A|\times \left|\int_{a}^{b}f(x)^2dx\right| $$ holds? Namely, can we find something about A such that the inequality $|\int_{a}^{b}f(x)dx|^2\ge |A|\times |\int_{a}^{b}f(x)^2dx| $ holds???
Anyone can help me? Thanks!
 A: Your inequalities are reversed. If you reverse both, you get the following:
$$
\left|\int_a^bf(x)\,\mathrm{d}x\right|^2\le(b-a)\int_a^b|f(x)|^2\,\mathrm{d}x
$$
This inequality follows from Cauchy-Schwarz or Hölder's Inequality or Jensen's Inequality. The converse cannot hold for any $A$. Consider $f(x)=x$ on $[-1,1]$

Positive Functions
In the example above, we looked at a function whose positive and negative parts cancel, so that $\int_a^bf(x)\,\mathrm{d}x=0$ cannot bound $\int_a^bf(x)^2\,\mathrm{d}x\gt0$. However, if we restrict $f(x)\ge0$, we have that if $\int_a^bf(x)^2\,\mathrm{d}x\gt0$, then $\int_a^bf(x)\,\mathrm{d}x\gt0$. This removes the simple counterexample above.
However, consider $f_n(x)=nx^n$ on $[0,1]$. As $n\to\infty$,
$$
\left(\int_0^1f_n(x)\,\mathrm{d}x\right)^2=\left(\int_0^1nx^n\,\mathrm{d}x\right)^2=\frac{n^2}{(n+1)^2}\to1
$$
and
$$
\int_0^1f_n(x)^2\,\mathrm{d}x=\int_0^1n^2x^{2n}\,\mathrm{d}x=\frac{n^2}{2n+1}\to\infty
$$
Again, we have that there is no constant $A$ so that
$$
\int_0^1f_n(x)^2\,\mathrm{d}x\le A\left(\int_0^1f_n(x)\,\mathrm{d}x\right)^2
$$
since the right hand side is bounded by $A$, yet the left side can be made as large as possible.
A: No you can't,
because you can choose
$f(x)$ so that
$\int_a^b f(x) dx = 0$.
If you restrict
$f$ so that
it is always positive,
then I think you need
a bound on $f$.
(added later)
There is a reversed-inequality
of the discrete C-S inequality:
If $0 < m < a_i/b_i < M$,
then
$$\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)
\le
\frac{(M+m)^2}{4mM} \left(\sum_{i=1}^n a_i b_i\right)^2
$$
You can find the proof here:
http://www.artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality
Setting
$b_i = 1$
(to get an approximation for 
$\int_0^n$),
this becomes
If $0 < m < a_i < M$,
then
$$n\sum_{i=1}^n a_i^2
\le
\frac{(M+m)^2}{4mM} \left(\sum_{i=1}^n a_i \right)^2
$$
Setting
$a=0$, $b=n$,
and
$a_i=f(i)$,
so that
$\sum_{i=0}^n a_i
\approx \int_0^n f(x) dx
$,
this becomes
If $0 < m < f(x) < M$,
then
$$(b-a)\int_a^b f(x)^2 dx
\le
\frac{(M+m)^2}{4mM} \left(\int_a^b f(x) dx \right)^2
$$
Converting this
to mean value over the interval
by dividing by
$(b-a)^2$,
this inequality becomes
$$\frac1{b-a}\int_a^b f(x)^2 dx
\le
\frac{(M+m)^2}{4mM} \left(\frac1{b-a}\int_a^b f(x) dx \right)^2
$$
I'm sure this is well-known.
