I think the following is a simple combinatorial argument which constructs the most dominant semistandard $\lambda$-tableau of content $\mu$ whenever $\lambda\trianglerighteq\mu$. (n.b. I haven't followed the reference given in Richard Stanley's comment, so I don't know whether I'm duplicating what's done there.)
In a nutshell, the idea is "put the largest numbers as low as possible". So let $l$ be the length of $\mu$, and start the tableau by putting the $\mu_l$ $l$s in the bottom cells of columns as far to the left as possible subject to the condition that if any box has an $l$ and there is a box directly to the right, then that box must also have an $l$ in it. If we let $j$ be maximal such that $\lambda_j\geq\mu_l$, this means that we are putting $\lambda_x-\lambda_{x+1}$ $l$s at the end of row $x$ for each $x>j$, and $\mu_l-\lambda_{j+1}$ $l$s at the end of row $j$.
To fill in the rest of the tableau, we work recursively. Let $\hat\lambda$ denote the partition whose Young diagram comprises the cells that are still empty, and let $\hat\mu$ be the partition $(\mu_1,\dots,\mu_{l-1})$. Then as long as $\hat\lambda\trianglerighteq\hat\mu$, we can fill in the rest of the tableau with a semistandard $\hat\lambda$-tableau of content $\hat\mu$.
So we need to show that $\hat\lambda_1+\dots+\hat\lambda_x\geq\hat\mu_1+\dots+\hat\mu_x$ for every $x$. For $x<j$ or $x\geq l$ this is immediate from the fact that $\lambda\trianglerighteq\mu$, so take $j\leq x<l$. Observe that $$\lambda_1+\dots+\lambda_x=n-(\lambda_{x+1}+\dots+\lambda_l)\geqslant n-(l-x)\lambda_{x+1}$$while $$\mu_1+\dots+\mu_x=n-(\mu_{x+1}+\dots+\mu_l)\leqslant n-(l-x)\mu_l.$$ So $$(\hat\lambda_1+\dots+\hat\lambda_x)-(\hat\mu_1+\dots+\hat\mu_x)=(\lambda_1+\dots+\lambda_{x+1}-\mu_l)-(\mu_1+\dots+\mu_x)\geqslant(l-x-1)(\mu_l-\lambda_{x+1})\geqslant0$$ as required.