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

### 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.)

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 $23.1322$.