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:
$$H(\lambda) \le Q(\lambda),
$$
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 thought this might be an easy exercise. Now I don't.

UPDATE (July 8, 2015): Petrov's elegant proof gives a stronger result. In particular, it proves what I suggested above, that the *variance* of complementary hooks $(q_{ij})$ is smaller than that of the usual hooks $(h_{ij})$. To see this, take $\varphi = x^2$ and use $Var(X) = E[X^2] - E[X]^2$.

Note also, as explained in the comments, the proof shows the hook numbers majorize the complementary hooks, when both are ordered from largest to smallest. For the example above: $5 \ge 3,$ $5+3\ge 3+3$, etc. This is quite remarkable and perhaps even counterintuitive.

SECOND UPDATE: We just wrote a paper on the subject with a different proof.