We shall use the following inequality:
For 0≤c1≤⋯≤ct, 0≤d1≤⋯≤dt,
(c1+d1)(c2+d2)⋯(ct+dt)≤(c1+dt)(c2+dt−1)⋯(ct+d1).1◯
This is because, for each i,
(ci+di)(ct+1−i+dt+1−i)≤(ci+dt+1−i)(ct+1−i+di).
Multiplying the above inequalities for i=1,…,t gives
i=1∏t((ci+di)(ct+1−i+dt+1−i))≤i=1∏t((ci+dt+1−i)(ct+1−i+di)),
and taking the square root on both sides of the above inequality gives (1).
For the original problem, let Li,j=∑q=1jai,q, Ri,j=∑q=jnai,q,
Ui,j=p=1∑iap,j,Di,j=p=i∑map,j.
Then
Xi,j=Li,j+Ui,j−ai,j,Yi,j=Di,j+Ri,j−ai,j.
Apply (1) to derive
i=1∏mj=1∏nXi,j=i=1∏mj=1∏n(Li,j+(Ui,j−ai,j))≥i=1∏mj=1∏n(Li,n+1−j+(Ui,j−ai,j))(since {Li,j}j=1n increases while {Ui,j−ai,j}j=1n decreases)≥i=1∏mj=1∏n(Ri,j+(Ui,j−ai,j))(since Li,n+1−j≥Ri,j)=j=1∏ni=1∏m((Ri,j−ai,j)+Ui,j)≥j=1∏ni=1∏m((Ri,j−ai,j)+Um+1−i,j)(since {Ri,j−ai,j}i=1m decreases while {Ui,j}i=1m increases)
≥j=1∏ni=1∏m((Ri,j−ai,j)+Di,j)(since Um+1−i,j≥Di,j)=i=1∏mj=1∏nYi,j.