For some $p$ and positive numbers $a=(a_1,\dots,a_n)\in\mathbb{R}^{+}$, let $M_p(a)$ denote the generalised or Hölder mean, defined as \begin{align} M_p = \left(\frac1n\sum_{i=1}^n a_i^p \right)^{1/p}, && a_1,\dots,a_n\ge 0, p\in\mathbb{R}^+ \end{align} One of the central properties of the generalized means is that for $p\le q$, it holds $M_p(a)\le M_q(a)$, which only becomes equality when $a_1=\dots=a_n.$ I wonder if there is any relationship of the following form \begin{align} \left(\frac{M_4(a)}{M_2(a)} \right)^2 \le \frac{M_2(a)}{M_1(a)} \end{align} The main intuition behind this guess is that, in the other extreme end, when only a single value of the is non-zero $a_1>0,a_2=\dots=a_n=0,$ we have $M_4(a)/M_2(a) = n^{1/4},$ and $M_2(a)/M_1(a)=n^{1/2}$.
Informally, this inequality states that the multiplicative "growth" of $M_p(a)$ as a function of $p$ is bounded in the $p\in[1,4]$ range, regardless of the value of $a$. I believe (correct me if this is wrong) is equivalent to saying that $f(p):=\log(M_p(a))$ has constant-Lipschitz in $p,$ (constant not dependent on $a$).
The inequality holds the other way around. More generally we can show the following.
It follows that $$ 3 f\left(\frac 12 \right) = 3 f\left(\frac 23 \cdot \frac 14 + \frac 13 \cdot 1 \right) \le 2 f\left(\frac 14 \right) + f(1) $$ and therefore $$ M_2^3 \le M_4^2 \cdot M_1 $$ or, equivalently, $$ \frac{M_2}{M_1} \le \left( \frac{M_4}{M_2}\right)^2 \, . $$ The case that one or more of the $a_i$ are zero can be obtained by a limiting process.
Generally for $0 < p < q < r$ one obtains $$ M_q^{(r-p)q} \le M_p^{(r-q)p} \cdot M_r^{(q-p)r} $$ by using the convexity condition for $f$ at the points $1/r< 1/q < 1/p$.
It remains to show that $$ f(x) = x \log \left(\sum_{i=1}^n a_i^{1/x}\right) - x \log(n) $$ is convex. A straight-forward calculation of the second derivative gives $$ f''(x) = \frac{1}{x^3}\left[ \frac{\sum a_i^{1/x} (\log(a_i))^2}{\sum a_i^{1/x}} - \left(\frac{\sum a_i^{1/x} \log(a_i)}{\sum a_i^{1/x}} \right)^2 \right] $$ and $f''(x) \ge 0$ follows from the Cauchy-Schwarz inequality: $$\left(\sum a_i^{1/x} \log(a_i)\right)^2 \le \left( \sum a_i^{1/x} \right) \cdot \left( \sum a_i^{1/x}(\log(a_i)^2 \right) \, . $$
Addendum: A simpler solution: Hölder's inequality with $p=3/2$ and $q=3$ gives $$ \sum_{i=1}^n a_i^2 = \sum_{i=1}^n a_i^{2/3} a_i^{4/3} \le \left( \sum_{i=1}^n a_i\right)^{2/3} \left( \sum_{i=1}^n a_i^4\right)^{1/3} $$ and raising this to the $(3/2)$-th power gives again $$ M_2^3 \le M_1 \cdot M_4^2 \, . $$