Does this moment inequality hold for any probability measure on the positive real line?

Question

Problem statement

Let $P$ be a probability measure on the positive real line and assume all it's raw moments, $\mu_k = \mathbb{E}[x^k]$, $k=1,2,\dots$ exist and $\mu_k < \infty$ for all $k$. Let $\mu = \mu_1$ be the mean of $P$ and assume $\mu > 0$.

Does the following inequality hold? $$ 4 \frac{\mu_3}{\mu^3} + 6 \frac{\mu_2}{\mu^2} - 9\left(\frac{\mu_2}{\mu^2}\right)^2 - 1 \leq 0 $$

Further remarks

I have tried to find a counterexample experimentally and found the above holds over a wide variety of continuous distributions (i.e. Gamma and Beta distributions) and may be tight in the sense that there are distributions which are just below zero.

Pointers

From Stieltjes moment problem we know that for any measure on the positive real line we must have that

$$ \det\left(\left[\begin{array}{cc} 1 & \mu\\ \mu & \mu_2 \end{array} \right]\right) > 0, \qquad\textrm{and}\qquad \det\left(\left[\begin{array}{cc} \mu & \mu_2\\ \mu_2 & \mu_3 \end{array}\right]\right) > 0, $$ from which we can infer that $$ \frac{\mu_2}{\mu^2} > 1, \qquad\textrm{and}\qquad \frac{\mu_3}{\mu^3} > \left(\frac{\mu_2}{\mu^2}\right)^2. $$

(But I have not been able to show the above inequality based on this.)

Answer 1

It doesn't hold. Note that if one chooses $P= \frac{1}{n^2} \delta_n + \frac{n^2-1}{n^2} \delta_x$ with $x$ chosen such that $\mu = 1$, i.e. $x = \frac{n^2-n}{n^2-1}$, then $\mu_3 \sim n$ asymptotically, while all other terms are bounded.

Or for a concrete counter example, take $n = 10$, then $\mu_2 = 1.8182$, $\mu_3 = 10.7438$ and the left hand side evaluates to $23.1322$.