In a book I found the following statement:
Let $\varphi(x)$ and $\psi(x)$ be square integrable, then $|\varphi \psi| \leq \frac{1}{2} |\varphi^2 + \psi^2|$. This implies, that every square integrabe function is absolutely integrable. To see this, let $\psi(x) = 1$
Now, I tried to proof the second sentence, and I'm not really sure that I did it correct.
Let $\varphi \in L_{2 \pi}^2$. Then we know that
\begin{align} \frac{1}{2\pi} \int_{0}^{2\pi} |\varphi(x)|^2 dx < \infty \end{align}
Let $\psi(x) = 1$, then (by the inequality mentioned above)
$$|\varphi| \leq \frac{1}{2} (\varphi^2 + 1) $$
Thus
$$\frac{1}{2 \pi} \int_{0}^{2 \pi} |\varphi(x)|dx \leq \frac{1}{2 \pi}\int_{0}^{2 \pi} (\varphi(x)^2 + 1) dx= \frac{1}{2 \pi}\int_{0}^{2 \pi} \varphi(x)^2dx + \frac{1}{2 \pi}\int_{0}^{2 \pi} dx < \infty$$
Is this correct? Thank you!
P.S. where does $|\varphi \psi| \leq 0.5 | \varphi^2 + \psi^2|$ come from?