Bounding $\sum_{k=-N}^{N} |a_k|$ where $a_k=\hat{f}(k)$ and $P_N(x)$ is a trigonometric polynomial of degree N

I need to prove that inequality above:

My approach was to write $|a_k| = |a_k| \cdot k \cdot \frac{1}{k}$ for all $k \neq 0$. Then, using the QM-GM inequality, we obtain:

$$|a_k| \leq \frac{1}{2}\left( |a_k|^2 k^2 + \frac{1}{k^2} \right).$$

Now, we know that the absolute values of the Fourier coefficients of $P'_N(x)$ are $|a_k|^2 k^2$, so we have:

$$\frac{1}{2} \sum_{k=-N}^{N} |a_k|^2 k^2 = \frac{1}{4\pi} \int_0^{2\pi} |P'_N(x)|^2 \, dx.$$

We can write this since we have Parseval's identity for the finite sum. also note that the $0$'s fourier coefficient of the derivative is $0$ by

and $P_N(x)$ is certainly continuous. So I did manage to work out the second integral (the derivative one) but how can I attack the first one? and get rid of the sum of $\frac{1}{k^2}$any ideas would be very much appreciated!