This is Exercise $4.1.2.$ in Katznelson's An Introduction to Harmonic Analysis.
Let $\{a_n\}_{n\in \Bbb Z}$ be a sequence of numbers and $\lambda > 0$. Find $\min \sum_{n\in \Bbb Z} |a_n|$ under the conditions $\sum_{n\in \Bbb Z} |a_n|^2 = 1$ and $\sum_{n\in \Bbb Z} |a_n|^4 = \lambda$.
This optimization problem seems to have no apparent connection with Fourier Analysis, and I'm just looking for a hint to get started. Of course, I've thought about $\{a_n\}$ being the Fourier coefficients of some function $f\in L^2(\Bbb T)$, i.e., $\widehat f(n) = a_n$ for all $n\in \Bbb Z$, $\|f\|_2 = \sum_{n\in \Bbb Z} |a_n|^2 = 1$, and $\sum_{n\in \Bbb Z} |\widehat f(n)|^4 = \lambda$. I'm stuck here.
Thanks!
Partial answer. In this answer, we prove the following three things: $$0<\lambda\leq 1,\tag{1}$$ $$\sum_{n\in\mathbb Z}|a_n|\geq \lambda^{-1/2},\tag{2}$$ and $$\text{the inequality $(2)$ is sharp for } \lambda\in\Lambda:=\left\{\frac1k:k=1,2,3,\cdots\right\}.\tag{3}$$
Proof of $(1)$. Since $\sum_{n\in \Bbb Z} |a_n|^2 = 1$, we have $|a_n|\in[0,1]$ for all $n\in\mathbb Z$, hence $|a_n|^4\leq|a_n|^2$ for all $n\in\mathbb Z$ and then $\lambda=\sum_{n\in \Bbb Z} |a_n|^4\leq \sum_{n\in \Bbb Z} |a_n|^2 = 1$.
Proof of $(2)$. It follows from Hölder's inequality that \begin{align*} 1=\sum_{n\in \Bbb Z} |a_n|^2&=\sum_{n\in \Bbb Z} |a_n|^{\frac23}|a_n|^{\frac43}\leq \left(\sum_{n\in \Bbb Z}|a_n|^{\frac23\cdot\frac32}\right)^{\frac23}\left(\sum_{n\in \Bbb Z}|a_n|^{\frac43\cdot3}\right)^{\frac13}\\ &=\left(\sum_{n\in \Bbb Z}|a_n|\right)^{\frac23}\left(\sum_{n\in \Bbb Z}|a_n|^{4}\right)^{\frac13}=\left(\sum_{n\in \Bbb Z}|a_n|\right)^{\frac23}\lambda^{\frac13}, \end{align*} hence $(2)$ holds.
Proof of $(3)$. If $\lambda=\frac1k$ for some $k\in\mathbb N_{\geq1}$, we consider the sequence $\{a_n\}_{n\in\mathbb Z}$ defined by $$a_n=\begin{cases} \sqrt\lambda, & 1\leq n\leq k,\\ 0, & \text{otherwise};\end{cases}$$ then $$\sum_{n\in\mathbb Z}|a_n|^2=k\lambda=1,\qquad \sum_{n\in\mathbb Z}|a_n|^4=k\lambda^2=\lambda,$$ and $\sum_{n\in\mathbb Z}|a_n|=k\sqrt\lambda=\lambda^{-1/2}$.
I don't know if $(2)$ is sharp for $\lambda\in(0,1]\setminus\Lambda$.