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][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 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][3] with a different proof.  


  [1]: http://www.math.ucla.edu/~pak/papers/excited.pdf
  [2]: https://math.stackexchange.com/questions/1851337/inequality-for-hook-numbers-in-young-diagrams
  [3]: http://www.math.ucla.edu/~pak/papers/Hook-ineq2.pdf