TITLE: Is there a short proof that the Kostka number $K_{\lambda \mu}$ is non-zero whenever $\lambda$ dominates $\mu$?
QUESTION [22 upvotes]: This is maybe a little basic for MathOverflow, but I'm hoping it will get some interesting answers.
Let $\unrhd$ be the dominance order on partitions of $n \in \mathbb{N}$.
For partitions $\lambda$ and $\mu$ of $n$, the Kostka Number $K_{\lambda\mu}$ is the number of semistandard Young tableaux of shape $\lambda$ and content $\mu$.
If $t$ is such a tableau then the $\mu_1+\cdots +\mu_r$ entries $k$ of $t$ such that $k \le r$ all lie in the first $r$ rows of $t$. Hence $\lambda_1 + \cdots + \lambda_r \ge \mu_1 + \cdots + \mu_r$ for each $r$ and so $\lambda \unrhd \mu$.

Is there a short combinatorial proof of the converse: if $\lambda \unrhd \mu$ then $K_{\lambda\mu} > 0$?

A constructive proof, maybe using the characterization of neighbours in the dominance order by single box moves on Young diagrams, would be especially welcome.
Edit. Using some representation theory it's possible to make this strategy work. The following is the symmetric functions version of Theorem 2.2.20 in the textbook by James and Kerber. Obviously $K_{\lambda\lambda} = 1$. Let $\mu$ and $\mu^\star$ be neighbours 
in the dominance order, with $\mu \rhd \mu^\star$, so $\mu^\star_i=\mu_i-1$, $\mu^\star_j = \mu_j+1$ for some $i < j$, and $\mu^\star_k = \mu_k$ if $k\not=i,j$. Since $h_{(a,b)} = h_{(a+1,b-1)} + s_{(a,b)}$ whenever $a \ge b$, we have
$$h_{\mu^\star} = \bigl(\prod_{k\not=i,j} h_{\mu_k} \bigr) h_{\mu_i-1}h_{\mu_j+1}=
\bigl(\prod_{k\not=i,j}h_{\mu_k}\bigr) \bigl( h_{\mu_i}h_{\mu_j} + s_{(\mu_i-1,\mu_j+1)} \bigr) $$
and so if $f = \prod_{k\not=i,j}h_{\mu_k}$ then
$$K_{\lambda\mu^\star} = \langle s_\lambda, h_{\mu^\star} \rangle = \langle s_\lambda, h_\mu \rangle + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle = K_{\lambda\mu} + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle \ge K_{\lambda\mu}.$$

REPLY [3 votes]: The following argument is reproduced from the proof of Lemma 3.7.3 in the book Representation Theory of the Symmetric Groups: The Okounkov–Vershik Approach, Character Formulas, and the Partition Algebras, by Tullio Ceccherini-Silberstein, Fabio Scarabotti, and Filippo Tolli (Cambridge University Press, 2010).
The proof is by induction on $n$. Suppose that $\lambda$ has $k$ parts and $\mu$ has $h$ parts, and that $\lambda\ge \mu$. Say that the positive integer $i$ is removable for the pair $(\lambda,\mu)$ if,
after setting
$$\tilde\lambda := (\lambda_1, \lambda_2, \ldots, \lambda_{i-1}, \lambda_i-1, \lambda_{i+1}, \ldots, \lambda_k)$$
and
$$\tilde\mu := (\mu_1, \mu_2, \ldots, \mu_{h-1}, \mu_h-1),$$
$\tilde\lambda$ is still a partition (i.e., $\lambda_i > \lambda_{i+1}$) and $\tilde\lambda \ge \tilde\mu$.  In particular, if the $i$th row of a SSYT of shape $\lambda$ and content $\mu$ contains the number $h$, then $i$ is removable for $(\lambda,\mu)$. Also, $k$ is always removable for the pair $(\lambda,\mu)$.
Let $i$ be the smallest removable integer for the pair $(\lambda,\mu)$, and let $\tilde\lambda$ and $\tilde\mu$ be as above. By the inductive hypothesis, we may construct a SSYT $\tilde T$ of shape $\tilde\lambda$ and content $\tilde\mu$. Moreover, by minimality of $i$, $\tilde T$ does not contain an entry $h$ in any row $j<i$. So if we add to $\tilde T$ a box at the end row $i$ with entry $h$, then we get a SSYT $T$ of shape $\lambda$ and content $\mu$.