We define the following inner product on intergrable, $2\pi$ periodic functions from $\mathbb{R}$ to $\mathbb{C}$:
$$\langle f,g\rangle = \frac{1}{2\pi} \int_{-\pi}^\pi f(t)\overline{g(t)}\ dt$$
I need to prove that:
$$\int_{-\pi}^\pi \left|f(t)\right| dt \le \sqrt{2\pi}\sqrt{\int_{-\pi}^\pi \left|f(t)\right|^2\ dt} = 2\pi \|f\|$$
Now, Cauchy-Schwarz inequality seemed to me perfect for this:
$$2\pi\|f\| =\|2\pi\|\|f\| \ge |\langle f, 2\pi \rangle| = \left| \int_{-\pi}^\pi f(t)\ dt\right|$$
but of course, we need the absolute value for the integrand.
I've also tried to use the fact that $f$ is bounded but that didn't yield anything useful.
Your are almost done, just note that $$ \def\norm#1{\left\|#1\right\|}\def\abs#1{\left|#1\right|}\norm{\abs f} = \norm{f}, $$ now start your calculations with $$ 2\pi\norm f = \cdots = \abs{\left<\abs f, 2\pi\right>} = \int_{-\pi}^\pi \abs{f(t)}\, dt $$ and you are done.