Problem:
The 2010 positive numbers a1​,a2​,…,a2010​ satisfy the inequality ai​aj​≤i+j for all distinct indices i,j. Determine, with proof, the largest possible value of the product a1​a2​⋯a2010​.
Solution:
Solution from Gabriel Carroll: Multiplying together the inequalities a2i−1​a2i​≤4i−1 for i=1,2,…,1005, we get
a1​a2​⋯a2010​≤3⋅7⋅11⋯4019(1)
The tricky part is to show that this bound can be attained.
and define ai​ for i<2008 by downward induction using the recursion
ai​=(2i+1)/ai+1​
We then have
ai​aj​=i+j whenever j=i+1 or i=2008,j=2010(2)
We will show that (2) implies ai​aj​≤i+j for all i<j, so that this sequence satisfies the hypotheses of the problem. Since a2i−1​a2i​=4i−1 for i=1,…,1005, the inequality (1) is an equality, so the bound is attained.
We show that ai​aj​≤i+j for i<j by downward induction on i+j. There are several cases:
If j=i+1, or i=2008,j=2010, then ai​aj​=i+j, from (2).
The first inequality holds by applying the induction hypothesis for (i+2,i+4), and (2) for the other pairs. The second inequality can again be checked by multiplying out: (2i+1)(2i+5)(2i+6)=8i3+48i2+82i+30<8i3+48i2+82i+42=(2i+2)(2i+3)(2i+7).
Here we have used the induction hypothesis for (i+2,j), and again we check the last inequality by multiplying out: (2i+1)(i+2+j)=2i2+5i+2+2ij+j<2i2+3i+2ij+3j=(2i+3)(i+j).
This covers all the cases and shows that ai​aj​≤i+j for all i<j, as required.
Variant Solution by Paul Zeitz: It is possible to come up with a semi-alternative solution, after constructing the sequence, by observing that when the two indices differ by an even number, you can divide out precisely. For example, if you wanted to look at a3​a8​, you would use the fact that a3​a4​a5​a6​a7​a8​=(7)(11)(15) and a4​a5​a6​a7​=(9)(13). Hence we need to check that (7)(11)(15)/((9)(13))<11, which is easy AMGM/ Symmetry.
However, this attractive method requires much more subtlety when the indices differ by an odd number. It can be pulled off, but now you need, as far as I know, either to use the precise value of a2010​ or establish inequalities for (ak​)2 for all values of k. It is ugly, but it may be attempted.