I am interested in an estimation for factorials, since I've been lately working with the dominance order of permutations. Let us say $\mu$ and $\lambda$ are partitions of a natural number $n$, that means $\mu=(\mu_1,\ldots, \mu_k)$ where $\mu_{\ell} \geq \mu_{\ell+1}$ for all $1 \leq \ell \leq k-1$ and $\mu_1+ \ldots + \mu_k=n$. Now I have that $\mu$ dominates $\lambda=(\lambda_1,\ldots, \lambda_{\ell})$, i.e.
$\mu_1 + \ldots \mu_m \geq \lambda_1 + \ldots+ \lambda_m$ for all $m$, where we define $\mu_m=0$ ($\lambda_m=0$) if $m\geq k$ ($m \geq \ell$).
Is there a proof to show that under these conditions the estimation
$\prod_{i=1}^m \mu_i! \geq \prod_{i=1}^m \lambda_i!$ holds for each $m$?
I was able to prove it in the case $m=2$ by showing that $\frac{\mu_1! \mu_2!}{\lambda_1! \lambda_2!} \geq 1$ holds, where I have essentially used that $\mu_1 \geq \lambda_1$ is true. In the general case I do not know, if $\mu_k \geq \lambda_k$ for an arbitrary $k$, therefore I do not know how to prove it, since induction does not work in this way.
Do you have any idea?
P.s. the proof for $m=2$:
If $\mu_2 \geq \lambda_2$ there is nothing to show, thus we suppose that $\mu_2 \leq \lambda_2$, which implies
$0 \leq \lambda_2-\mu_2 \leq \mu_1 - \lambda_1$,
where the last estimation holds by using the assumption $\mu_1+ \mu_2 \geq \lambda_1 + \lambda_2$. But with this we have
$\frac{\mu_1! \mu_2!}{\lambda_1!\lambda_2!} = \frac{\mu_1 \cdots (\lambda_1+1)\mu_2!}{\lambda_2!}=\frac{\mu_1 \cdots (\lambda_1+1)}{\lambda_2\cdots (\mu_2+1)}$.
But since by assumtion there are less factors in the denominator than in the numerator and $\mu_1 \geq \lambda_1 \geq \lambda_2$ holds, we obtain $\frac{\mu_1! \mu_2!}{\lambda_1!\lambda_2!} \geq 1$.