For a partition $\lambda \vdash n$, define $\dim \lambda$ to be the number of standard Young tableaux of shape $\lambda$, and $\dim \lambda/\(k)$ as the number of standard Young tableaux with $1,2,\ldots,k$ in the first row. It is not hard to see that $\dim \lambda/(k)$ is equal to the number of skew standard tableaux of shape $\lambda/(k)$ if $k \leq \lambda_1$ and $0$ if $k > \lambda_1$.

We write $\lambda \unrhd \beta$ to mean that $\beta$ precedes $\lambda$ in the standard dominance ordering of partitions.

I have an inelegant proof of the following statement: For every $\lambda \vdash n$ and $\beta \vdash n$ where $\lambda \unrhd \beta$, and any positive integer $k$, we have

$$ \frac{\dim \lambda / (k)}{\dim \lambda} \geq \frac{\dim \beta / (k)}{\dim \beta}. $$

This seems to me like a statement that should be in the literature, yet I could not find it. Does anyone have a reference for (or a clean proof of) this statement?