2010 USAMO Problems/Problem 3

Problem

The $2010$ positive numbers $a_1, a_2, \ldots , a_{2010}$ satisfy the inequality $a_ia_j \le i+j$ for all distinct indices $i, j$ . Determine, with proof, the largest possible value of the product $a_1a_2\cdots a_{2010}$ .

Solution

The largest possible value is $\prod_{i=1}^{1005}(4i-1) = 3\times 7 \times \ldots \times 4019.$

Proof

No larger value is possible, since for each consecutive pair of elements: $(a_{2i-1},a_{2i}), 1\le i \le 1005$ , the product is at most $(2i-1) + 2i = 4i - 1$ , and so the product of all the pairs is at most:

$\prod_{i=1}^{1005}(4i-1).$

If we can demonstrate a sequence in which for all $1 \le i \le 1005$ the product $a_{2i-1}a_{2i} = 4i-1$ , and all the inequalities are satisfied, the above upper bound will be achieved and the proof complete.

We will construct sequences of an arbitrarily large even length $2n \ge 4$ , in which:

$\begin{cases} a_ia_j = i+j = 2i+1 & j=i+1, \\ a_ia_j = i+j = 2n-2 & j = i+2 = 2n, \\ a_ia_j < i+j & i \ne 2n-2, \mbox{ and } i < j-1. \end{cases}$

Given $a_1$ , from the equations $a_ia_{i+1} = 2i+1,\; 1\le i\le 2n-1$ , we obtain the whole sequence recursively: $a_1 = a_1,\; a_2 = 3/a_1,\; a_3 = 5/a_2 = 5a_1/3,\; a_4 = 7/a_3 = (3\cdot 7)/(5a_1) \ldots.$ And as a result:

$a_{i+2} = a_i \cdot \frac{2i+3}{2i+1}\quad 1 \le i \le 2n-2.$

The same equations $a_ia_{i+1} = 2i+1$ can be used to compute the whole sequence from any other known term.

We will often need to compare fractions in which the numerator and denominator are both positive, with fractions in which a positive term is added to both. Suppose $p, q, r$ are three positive real numbers, then:

$\begin{align*} p &> q \implies & \frac{p+r}{q+r} &< \frac{p}{q} \\ p &< q \implies & \frac{p+r}{q+r} &> \frac{p}{q} \\ & \mbox{and in either case} & \frac{p+r}{q} &> \frac{p}{q} \end{align*}$

Returning to the problem in hand, for $i < j$ , $a_ia_j \le i+j \implies a_ia_{j+2} < i+j+2$ . If it were otherwise, we would have for some $i < j$ :

$\frac{a_{j+2}}{a_j} \ge \frac{i+j+2}{i+j} > \frac{j+j+2}{j+j} = \frac{2j+2}{2j},$

but the ratio of the $j^{\mathrm{th}}$ term to the next term of the same parity is $(2j+3)/(2j+1) < (2j+2)/2j$ , so our assumption is impossible. Note, we are using the fact that if $p > q > 0$ and $r > 0$ it follows that $(p+r)/(q+r) < p/q$

Therefore, we need only verify inequalities with an index difference of $1$ or $2$ , as these imply the rest.

Now, when the indices differ by $1$ we have ensured equality (and hence the desired inequalities) by construction. So, we only need to prove the inequalities for successive even index and successive odd index pairs, i.e. for every index $i > 2$ , prove $a_{i-2}a_i \le 2i-2$ .

We now compare $a_ia_{i+2}/(2i+2)$ with $a_{i+2}a_{i+4}/(2i+6)$ . By our recurrence relations:

$\begin{align*} \frac{a_{i+2}a_{i+4}}{2i+6} &= \frac{a_ia_{i+2}}{2i+2} \cdot \frac{2i+2}{2i+6} \cdot \frac{a_{i+4}}{a_{i+2}}\cdot \frac{a_{i+2}}{a_i} \\ &= \frac{a_ia_{i+2}}{2i+2}\cdot \frac{i+1}{i+3}\cdot \frac{2i+7}{2i+5}\cdot \frac{2i+3}{2i+1} \\ &= \frac{a_ia_{i+2}}{2i+2}\cdot \frac{(i+1)(2i+7)(2i+3)}{(i+3)(2i+5)(2i+1)} \\ &= \frac{a_ia_{i+2}}{2i+2}\cdot \frac{4i^3+24i^2+41i+21}{4i^3+24i^2+41i+15} \\ &= \frac{a_ia_{i+2}}{2i+2}\cdot \left(1+\frac{6}{4i^3+24i^2+41i+15}\right) \\ &> \frac{a_ia_{i+2}}{2i+2}. \end{align*}$

So, for both odd and even index pairs, the strict inequality $a_ia_{i+2} < 2i+2$ follows from $a_{i+2}a_{i+4} \le 2i+6$ and we need only prove the inequalities $a_{2n-3}a_{2n-1} \le 4n-4$ and $a_{2n-2}a_{2n} \le 4n-2$ , the second of which holds (as an equality) by construction, so only the first remains.

We have not yet used the equation $a_{2n-2}a_{2n} = 4n-2$ , with this we can solve for the last three terms (or equivalently their squares) and thus compute the whole sequence. From the equations:

$\begin{align*} a_{2n-1}a_{2n} &= 4n-1 \\ a_{2n-2}a_{2n} &= 4n-2 \\ a_{2n-2}a_{2n-1} &= 4n-3 \end{align*}$

multiplying any two and dividing by the third, we get:

$\begin{align*} a_{2n}^2 &= \frac{(4n-2)(4n-1)}{4n-3} \\ a_{2n-1}^2 &= \frac{(4n-3)(4n-1)}{4n-2} \\ a_{2n-2}^2 &= \frac{(4n-3)(4n-2)}{4n-1} \end{align*}$

from which,

$a_{2n-3}^2 = \frac{(4n-5)^2}{a_{2n-2}^2} = \frac{(4n-5)^2(4n-1)}{(4n-3)(4n-2)}$

With the squares of the last four terms in hand, we can now verify the only non-redundant inequality:

$\begin{align*} a_{2n-3}^2a_{2n-1}^2 &= \frac{(4n-5)^2}{a_{2n-2}^2}a_{2n-1}^2 \\ &= \left(\frac{(4n-5)(4n-1)}{4n-2}\right)^2 \\ &= \left(\frac{16n^2 -24n + 5}{4n-2}\right)^2 \\ &< \left(\frac{16n^2 -24n + 8}{4n-2}\right)^2 \\ &= (4n-4)^2. \end{align*}$

The inequality above follows because the numerator and denominator are both positive for $n > 1$ .

This completes the construction and the proof of all the inequalities, which miraculously reduced to just one inequality for the last pair of odd indices.

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2010 USAMO Problems/Problem 3

Problem

Solution

Proof