Maximizing the ratio $\|p\|_\infty/\|p\|_2^2$ for a probability distribution $p$ on $n$ elements Let $p$ be an arbitrary probability distribution on $[n]:= \{1,2,\dots,n\}$, for some $n\geq 2$. Equating $p$ with its probability mass function, it is not hard to show that its $\ell_r$ norms satisfy
$$
\|p\|_2^2 \leq \|p\|_\infty \leq \|p\|_2 \tag{1}
$$
and in particular
$$
1 \leq \frac{\|p\|_\infty}{\|p\|_2^2} \leq \sqrt{n} \tag{2}
$$
Further, the lower bound is tight in the worst case, as shown by taking $p$ to be the uniform distribution on $[n]$. I am trying to find for which extremal distribution $p$ the upper bound is tight. (Or asymptotically tight, considering a sequence $(p^{(n)})_{n\geq 2}$ of distributions, and letting $n\to \infty$.)
I am able to find a sequence of distributions $(p^{(n)})_{n\geq 2}$ such that
$$
\frac{\|p^{(n)}\|_\infty}{\|p^{(n)}\|_2^2} \operatorname*{\sim}_{n\to\infty} \frac{2\sqrt{n}}{\ln n} \tag{3}
$$
which is almost tight, up to a factor $\frac{2}{\ln n}$:
$$
p^{(n)}_k = \frac{1}{\sum_{\ell=1}^n \frac{1}{\sqrt{\ell}}}\cdot \frac{1}{\sqrt{k}}, \qquad 1\leq k\leq n
$$
as then
$
\frac{\|p^{(n)}\|_\infty}{\|p^{(n)}\|_2^2}  = \frac{\sum_{\ell=1}^n \frac{1}{\sqrt{\ell}}}{H_n}
$, giving (3).

For every $n\geq 2$, is there a distribution $p$ on $[n]$ achieving the upper bound of (2)? What about in the asymptotic sense above, with a sequence $(p^{(n)})_{n\geq 2}$? If not, can the upper bound in (2) be improved by a logarithmic factor (i.e., is (3) tight)?

 A: Update/clarification. (with thanks to Nathan's pointing this out!) My original answer below was not a proof that $\|p^*\|_\infty/\|p^*\|_2^2\le2\sqrt n/5$. This is because the minimum I obtained was the minimum for the function $\sqrt n\|p\|_2^2-\|p\|_\infty$, so a direct substitution would lead to a misleading result of the desired ratio, as we have $\|p\|_2^2$ on both sides.
With this in mind, the optimisation problem \begin{align}\min&\quad p_n\left(\sum_{i=1}^np_i^2\right)^{-1}\\\text{s.t.}&\quad\sum_{i=1}^np_i=1\end{align} yields $p_1=\cdots=p_{n-1}$ as $\partial/\partial p_j$ yields the same value for all $1\le j\le n-1$. As it turns out, we do not need to use the quartic equality obtained through $\partial/\partial p_n$, so the p.m.f. constraint $(n-1)p_k+p_n=1$ for any $1\le k\le n-1$ yields $$\frac{\|p\|_\infty}{\|p\|_2^2}\le\frac{(n-1)p_n}{(1-p_n)^2+(n-1)p_n^2}=\frac{(n-1)p_n}{np_n^2-2p_n+1}.$$ Its reciprocal $(np_n-2+1/p_n)/(n-1)$ is minimised at $p_n^*=1/\sqrt n$ so that $$\frac{\|p\|_\infty}{\|p\|_2^2}\le\frac{n-1}{\sqrt n(1-2/\sqrt n+1)}=\frac{n-1}{2(\sqrt n-1)}=\frac{\sqrt n+1}2.$$
More generally, we have $$\frac{\|p\|_\infty}{\|p\|_r^r}\le\frac{(n-1)^{r-1}s}{(1-s)^r+(n-1)^{r-1}s^r}$$ where $s$ solves the polynomial equation $$(1+(n-1)^{r-1})s^{r+1}+\sum_{q=2}^{r-1}(q-1)\binom rq(-s)^q-1=0.$$

Original answer. The upper bound in $(2)$ is equivalent to the inequality $$\frac{p_n}{p_1^2+\cdots+p_n^2}\le\sqrt n$$ where $p_1+\cdots+p_n=1$ and $p_n:=\max\{p_1,\cdots,p_n\}$ without loss of generality. We can formulate this optimisation problem as \begin{align}\min&\quad\sqrt n\sum_{i=1}^np_i^2-p_n\\\text{s.t.}&\quad\sum_{i=1}^np_i=1\end{align} so consider the function $$f({\bf p})=\sqrt n\sum_{i=1}^np_i^2-p_n-\lambda\left(\sum_{i=1}^np_i-1\right)$$ where $\lambda$ is a Lagrange multiplier. We have $\partial f/\partial p_j=2\sqrt np_j-\lambda$ for all $j\le n-1$ and $\partial f/\partial p_n=2\sqrt np_n-1-\lambda$. Setting $\nabla f({\bf p})=0$ yields $$p_j=\frac\lambda{2\sqrt n},\quad p_n=\frac{\lambda+1}{2\sqrt n},\quad\lambda=\frac{2\sqrt n-1}n$$ using the p.m.f. constraint. The objective function thus has a minimum at $$\sqrt n\left((n-1)\frac{(2\sqrt n-1)^2}{4n^3}+\frac{(2\sqrt n+n-1)^2}{4n^3}\right)-\frac{2\sqrt n+n-1}{2n\sqrt n}$$ which simplifies to $$\frac{(\sqrt n-1)(3\sqrt n-1)}{4n\sqrt n}\le\sqrt n\sum_{i=1}^np_i^2-p_n$$ so $$\frac{\|p\|_\infty}{\|p\|_2^2}\le\sqrt n-\frac{(\sqrt n-1)(3\sqrt n-1)}{4n\sqrt n\|p\|_2^2}.$$ This means that for $n\ge2$, the inequality $\|p\|_\infty/\|p\|_2^2\le\sqrt n$ must be strict.
At the minimum ${\bf p}^*=\begin{pmatrix}\frac{2\sqrt n-1}{2n\sqrt n}&\cdots&\frac{2\sqrt n-1}{2n\sqrt n}&\frac{2\sqrt n+n-1}{2n\sqrt n}\end{pmatrix}^\top$, we have $$\|p^*\|_2^2=\frac{5n-1}{4n^2}=\frac5{4n}+o\left(\frac1n\right)$$ so the ratio $\|p^*\|_\infty/\|p^*\|_2^2$ is asymptotically $$(1+o(1))\sqrt n\left(1-\frac3{4\cdot5/4}\right)=(1+o(1))\frac25\sqrt n.$$
A: I think the answer by TheSimpliFire was only almost correct but not quite because it identified the wrong $p$ on the $(\|p\|_\infty,-\|p\|_2^2)$ Pareto frontier. Here's my attempt, which shows that $\lim_{n\to\infty}\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\sqrt{n}\|p\|_2^2}=\frac12$, approaching it from above.
Let $\Delta^n$ denote the $(n-1)$-dimensional simplex. We are interested in $\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\|p\|_2^2}$. The solution to this lives somewhere on the $(\|p\|_\infty,-\|p\|_2^2)$ Pareto frontier, which consists of solutions to $\max_{p\in\Delta^n}(\|p\|_\infty-\mu\|p\|_2^2)$ for some $\mu\in[0,\infty]$. The solutions to the latter are the same as permutations of solutions to $\max_{p\in\Delta^n}(p_1-\mu\|p\|_2^2)$ as TheSimpliFire noted. This later optimization problem is convex (i.e., max of concave objective over convex set) and permutation symmetric in entries $2$ through $n$. So an optimal $p$ must have the form $p_i=(1-p_1)/(n-1)~\forall i\geq2$ such that $p_1\geq 1/n$. Such $p$ has $\frac{\|p\|_\infty}{\|p\|_2^2}=\frac{p_1}{p_1^2+(n-1)(\frac{1-p_1}{n-1})^2}$. Now it's just some univariate calculus (aka Wolfram Mathematica) to get that
$$\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\|p\|_2^2}=\max_{p_1\in[1/n,1]}\frac{p_1}{p_1^2+(n-1)(\frac{1-p_1}{n-1})^2}=\frac{\sqrt{n}+1}{2}.$$
Hence,
$\max_{p\in\Delta^n}\frac{\|p\|_\infty}{\sqrt{n}\|p\|_2^2}$ is both $\geq1/2$ and converges $\to1/2$ as $n\to\infty$.
