## An inequality for the ratio of standard Young tableau with {1,2,…,k} in the first row

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?

-
 "This seems to me like a statement that should be in the literature". I can't say I have seen too many inequalities about Young tableaux in literature... – darij grinberg Jan 12 at 2:19 I agree with darij grinberg's comment. Do you know if there is a refinement to a chain of results: $\frac{\dim \lambda / (k)}{\dim \beta / (k)} \ge \frac{\dim \lambda / (k-1)}{\dim \beta / (k-1)}$ for $k \ge 1$? – Mark Wildon Jan 12 at 19:36 Thanks, darij and Mark! I haven't seen too many inequalities either, but this is most definitely not my area of expertise. Mark, I did not consider the refinement you propose, so I have no idea if it holds or not (though my intuition says that it does). – KEW Jan 12 at 21:37