integrals inequalities $$
\left( {\int\limits_0^1 {f^2(x)\ \text{d}x} }\right)^{\frac{1}
{2}} \  \geqslant \quad \int\limits_0^1 {\left| {f(x)} \right|\ \text{d}x} 
$$
I can't prove it )=

Define
$$
\mu = \int_0^1 {|f(x)|\,dx} 
$$
and
$$
\sigma^2 = \int_0^1 {(|f(x)| - \mu )^2 \,dx} .
$$
Then
$$
\sigma^2 = \int_0^1 {f^2 (x)\,dx}  - 2\mu \int_0^1 {|f(x)|\,dx}  + \mu ^2  = \int_0^1 {f^2 (x)\,dx}  - \mu ^2 .
$$
Since $\sigma^2 \geq 0$, 
$$
\int_0^1 {f^2 (x)\,dx}  \geq \mu ^2.
$$
Taking square roots of both sides yields the desired result:
$$
\bigg(\int_0^1 {f^2 (x)\,dx} \bigg)^{1/2}  \ge \int_0^1 {|f(x)|\,dx}. 
$$
EDIT: The idea used here is that for a random variable $Y$ with finite second moment,
$$
{\rm Var}(Y) := {\rm E}[Y - {\rm E}(Y)]^2  \geq 0.
$$
So,
$$
{\rm Var}(Y) = {\rm E}(Y^2) - 2{\rm E}(Y){\rm E}(Y)+  {\rm E}^2 (Y) =  {\rm E}(Y^2) - {\rm E}^2 (Y),
$$
and hence
$$
{\rm E}(Y^2) \geq {\rm E}^2 (Y).
$$
To relate this to the question at hand, let $X$ be a uniform$[0,1]$ random variable, and let $Y=|f(X)|$ (for $f$ a square-integrable function on $[0,1]$).
Then
$$
{\rm E}(Y^2) = {\rm E}[f^2 (X)] = \int_0^1 {f^2 (x)\,dx} 
$$
and
$$
{\rm E}^2 (Y) = {\rm E}^2 (|f(X)|) = \bigg(\int_0^1 {|f(x)|\,dx} \bigg)^2 .
$$
Hence
$$
\int_0^1 {f^2 (x)\,dx}  \geq \bigg(\int_0^1 {|f(x)|\,dx} \bigg)^2 ,
$$
which gives the desired result after taking square roots.
