relationship between generalized mean ratios $M_4(a)/M_2(a)$ vs $M_2(a)/M_1(a)$ 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 $a$).
 A: The inequality holds the other way around. More generally we can show the following.

For fixed positive numbers $a_1, \ldots, a_n$  and $x > 0$ is the function
$$
 f(x) = \log M_{1/x}(a_1, \ldots, a_n)
$$
convex.

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
$$
 M_2^3 \le M_1 \cdot M_4^2 \, .
$$
