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. The proof of the main result is algebraic and quite involved.

P.S. Originally posted on MSE since I originally thought this might be an exercise. Now I don't.