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 )=

$$\int_0^1 |f(x)| \, dx = \int_0^1 |1||f(x)| \, dx \leq \sqrt{\int_0^1 1 \, dx} \sqrt{\int_0^1 |f(x)|^2 \, dx} = \sqrt{\int_0^1 |f(x)|^2 \, dx}$$
By Cauchy-Schwarz.
