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](http://www.math.ucla.edu/~pak/papers/excited.pdf). The proof of the main result is algebraic and quite involved. 


P.S. Originally [posted](https://math.stackexchange.com/questions/1851337/inequality-for-hook-numbers-in-young-diagrams) 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](http://www.math.ucla.edu/~pak/papers/Hook-ineq2.pdf) with a different proof. 



Inequality for hook numbers in Young diagrams
Not sure, please check carefully. (Well, now more sure and the argument is more direct.)


I claim that the array $(h)$ majorates the array $(q)$, that is, 
$\sum \varphi (h\_{ij})\geqslant \sum \varphi(q\_{ij})$ for any convex function $\varphi$,
in particular for $-\log$, that is your inequality. 


Denote the hook lengths of the first (largest) column 
by $0<c\_1<c\_2<\dots<c\_m$. Then the hooks in $i$-th row, which
contains $c\_i-i+1$ squares, are all numbers from 1 to $c\_i$
except $c\_i-c\_1$, $c\_i-c\_2$, $\dots$, $c\_i-c\_{i-1}$ (this
elementary claim is well-known, it shows the equivalence of Frobenius and the
hook length formulae.) That is, 
$$
A:=\sum \varphi(h\_{ij})=\sum\_i \bigl(\varphi(1)+\ldots+\varphi(c\_i)\bigr)-\sum\_{i<j} \varphi(c\_j-c\_i).
$$
Clearly
$$
B:=\sum \varphi(q\_{ij})=\sum\_i \bigl(\varphi(m-i+1)+\ldots+\varphi(c\_i+m-2i+1)\bigr).
$$


For $i<j$, we have:
$$
\varphi(j-i)+\varphi(c\_j-i+1)\geqslant \varphi(c\_j-c\_i)+\varphi(c\_i+j-2i+1),
$$
since for $c\_j=c\_i+j-i$ the equality takes place, and the difference of two sides
increases as a function of $c\_j\in [c\_i+j-i,+\infty)$. 
Summing up these inequalities over all pairs $i<j$ we get the desired inequality $A-B\geqslant 0$.


If $\varphi$ is strictly convex, then all inequalities 
are sharp only for a rectangle. 


