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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, $$ A:=\sum \varphi(h_{ij})=\sum_i \left(\varphi(1)+\dots+\varphi(c_i)\right)-\sum_{i<j} \varphi(c_j-c_i). $$ Clearly $$ B:=\sum \varphi(q_{ij})=\sum_i \left(\varphi(m-i+1)+\dots+\varphi(c_i+m-2i+1)\right). $$

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 exactly the necessary inequality $A-B\geqslant 0$.

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

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.

Not sure, please check carefully.

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, $$ A:=\sum \varphi(h_{ij})=\sum_i \left(\varphi(1)+\dots+\varphi(c_i)\right)-\sum_{i<j} \varphi(c_j-c_i). $$ Clearly $$ B:=\sum \varphi(q_{ij})=\sum_i \left(\varphi(m-i+1)+\dots+\varphi(c_i+m-2i+1)\right). $$ Take the difference $A-B$ and consider it as a function of $c_m$ only (fix other $c_i$'s). It equals some constant (not depending on $c_m$) plus $$ \sum_{i=1}^{m-1} \varphi(c_m-i+1)-\varphi(c_m-c_i). $$ Each summand increases as a function of $c_m$, since $i-1<c_i$ and $\varphi$ is convex. That is, it is minimal for $c_m=c_{m-1}+1$. Well, so we may suppose that $c_m=c_{m-1}+1$ (two largest rows have equal length). Do this and consider the difference $A-B$ as a function of $c_{m-1}$. Again it is minimal for $c_{m-1}=c_{m-2}+1$, and so on.

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

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, $$ A:=\sum \varphi(h_{ij})=\sum_i \left(\varphi(1)+\dots+\varphi(c_i)\right)-\sum_{i<j} \varphi(c_j-c_i). $$ Clearly $$ B:=\sum \varphi(q_{ij})=\sum_i \left(\varphi(m-i+1)+\dots+\varphi(c_i+m-2i+1)\right). $$

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 exactly the necessary inequality $A-B\geqslant 0$.

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

Not sure, please check carefully.

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, $$ A:=\sum \varphi(h_{ij})=\sum_i \left(\varphi(1)+\dots+\varphi(c_i)\right)-\sum_{i<j} \varphi(c_j-c_i). $$ Clearly $$ B:=\sum \varphi(q_{ij})=\sum_i \left(\varphi(m-i+1)+\dots+\varphi(c_i+m-2i+1)\right). $$ Take the difference $A-B$ and consider it as a function of $c_m$ only (fix other $c_i$'s). It equals some constant (not depending on $c_m$) plus $$ \sum_{i=1}^{m-1} \varphi(c_m-i+1)-\varphi(c_m-c_i). $$ Each summand increases as a function of $c_m$, since $i-1<c_i$ and $\varphi$ is convex. That is, it is minimal for $c_m=c_{m-1}+1$. Well, so we may suppose that $c_m=c_{m-1}+1$ (two largest rows have equal length). Do this and consider the difference $A-B$ as a function of $c_{m-1}$. Again it is minimal for $c_{m-1}=c_{m-2}$, and so on.

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

Not sure, please check carefully.

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, $$ A:=\sum \varphi(h_{ij})=\sum_i \left(\varphi(1)+\dots+\varphi(c_i)\right)-\sum_{i<j} \varphi(c_j-c_i). $$ Clearly $$ B:=\sum \varphi(q_{ij})=\sum_i \left(\varphi(m-i+1)+\dots+\varphi(c_i+m-2i+1)\right). $$ Take the difference $A-B$ and consider it as a function of $c_m$ only (fix other $c_i$'s). It equals some constant (not depending on $c_m$) plus $$ \sum_{i=1}^{m-1} \varphi(c_m-i+1)-\varphi(c_m-c_i). $$ Each summand increases as a function of $c_m$, since $i-1<c_i$ and $\varphi$ is convex. That is, it is minimal for $c_m=c_{m-1}+1$. Well, so we may suppose that $c_m=c_{m-1}+1$ (two largest rows have equal length). Do this and consider the difference $A-B$ as a function of $c_{m-1}$. Again it is minimal for $c_{m-1}=c_{m-2}+1$, and so on.

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