$L^{2}$ spaces, bounding the measure of a set Let $f \in L^{2} ([0,1], \mathcal{A},\mu)$ where $\mu$ is the usual Lebesgue measure, such that $||f||_{2}=1$ and
$$
\int_{0}^{1} f d \mu \geq \alpha > 0
$$
Let $E_{\beta}= \{ x \in [0,1] | f(x) > \beta \}$.
Then, for every $\beta \in (0,\alpha)$
$$
\mu ( E_{\beta} ) \geq (\alpha-\beta)^{2}
$$
I had problems with this exercise. To be sincere, I don't know even where to start from. The only relevant inequality that I know is Chevyshev inequality but it bounds the measure of $E_{\beta}$ from above, not from below. I've tried to do some trick with integrals and charasteristic function to bound the $L^{2}$ norm of $\alpha-\beta$ from above with the $L^{2}$ norm of $\chi_{E_{\beta}}$, but I haven't succeded. My biggest concern is the lack of absolute values marks on the definition of $E_{\beta}$ which doesn't allow me to make the conclusion $x > y \Rightarrow x^{2} > y^{2}$.
Any hints?
 A: You said that you didn't know where to start.
Let's start with what we know, namely $\alpha \leq \int_0^1 f \, d \mu$. We now want to connect this to the quantity $\mu(E_\beta)$ that we want to show something about. The most natural approach is to write
$$
\alpha \leq \int_0^1 f \, d\mu = \int_0^1 f \, 1_{E_\beta} \, d \mu + \int_0^1 f \, 1_{E_\beta^c} \, d\mu.
$$
Now, note that $f \leq \beta$ on $E_\beta^c$, which implies that $\int_0^1 f \, 1_{E_\beta^c} \, d \mu \leq \int_0^1 \beta \, 1_{E_\beta^c} \, d \mu \leq \beta$, since $\beta \geq 0$.
We have thus shown
$$
\alpha \leq \int_0^1 f \, 1_{E_\beta} \, d \mu + \beta.
$$
First try to figure out yourself how to proceed from here. If you don't see it after staring at the equation for ~10 minutes, look at the hint below.

 Hint: Cauchy-Schwarz inequality.

And here the full solution:

 Note that until now we did not use the fact that $\|f\|_2 = 1$. To make use of this estimate, we use the Cauchy-Schwarz inequality, which shows
 \begin{align*}\alpha \leq \beta + \int_0^1 f \, 1_{E_\beta} \, d \mu \leq \beta + \int_0^1 |f| \, 1_{E_\beta} \, d \mu \leq \beta + \sqrt{\int_0^1 |f|^2 \, d \mu} \sqrt{\int_0^1 1_{E_\beta} \, d \mu} = \beta + \sqrt{\mu(E_\beta)}. \end{align*} Rearranging this inequality will give you the desired estimate.

