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 *complimentary 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: $$H(\lambda) \le Q(\lambda), $$ where the inequality becomes the equality only for rectangular shapes.* 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 than $h_{ij}$. **Note:** this inequality is a corollary of the results in [our recent paper][1]. The proof of the main result is algebraic and quite involved. P.S. Originally [posted][2] on MSE since I thought this might be an exercise. Now I don't. [1]: http://www.math.ucla.edu/~pak/papers/excited.pdf [2]: http://math.stackexchange.com/questions/1851337/inequality-for-hook-numbers-in-young-diagrams