Consider a Young diagram $\lambda = (\lambda_1,\ldots,\lambda_\ell)$. For a square $(i,j) \in \lambda$ define hook numbers $h_{ij} = \lambda_i + \lambda_j' -i - j +1$ and complementary hook numbers $q_{ij} = i + j -1$. Let $$H(\lambda) = \prod_{(i,j) \in \lambda} h_{ij}\,, \qquad Q(\lambda) = \prod_{(i,j) \in \lambda} q_{ij}\,. $$

Question: is there an elementary proof of the following inequality: $$H(\lambda) \le Q(\lambda). $$

For example, when $\lambda = (3,2,1)$ we have $$H(\lambda)=5\cdot 3 \cdot 3 \cdot 1\cdot 1 \cdot 1 = 45, \qquad Q(\lambda)=1\cdot 2 \cdot 2 \cdot 3\cdot 3 \cdot 3 = 108. $$ Let me mention that $$\sum_{(i,j) \in \lambda} h_{ij} = \sum_{(i,j) \in \lambda} q_{ij}, $$ so somehow this says that $q_{ij}$ are more evenly distributed.

Note: this inequality is a corollary of the main results in our paper. The proof of the main result is algebraic and quite involved.

UPDATE (7/7/2016) Cross-posted on MO where it was resolved.

  • 2
    $\begingroup$ I think belongs on mathoverflow.net. $\endgroup$ – Mosquite Jul 6 '16 at 20:24

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