I didn't qualify for the USAMO (index was about 30 points lower than the cutoff ), but I have a somewhat likely answer, albeit without any truly mathematical proof.

The thing is, the numbers can be non-integers as well.

I claimed that, for even $n$, the largest product of the reals $a_1$ through $a_n$ was simply the product of all positive integers equivalent to $3$ modulo $4$ up to $2n-1$. I want to hope that it's worth one point, but I'm not sure it's correct, and I'm not even sure it's nontrivial enough to be worth one point.

does nobody have a solution to this problem?

tjhance has a nice solution:

On another note, how many points do you think one would get if they constructed a sequence of the form $a_ia_{i+1}=2i+1$ AND used induction, but it was wrong?

I got a solution for #3 ! Sketch- I saw that you could choose different ways to multiply 1005 inequalities together (so that each $a_i$ appears once in the left hand sides to get an upper bound for the product. I claimed that the lowest such upper bound is the maximum value of the product (I'm not certain about this though...) So, $a_1a_2a_3\cdots a_{2010}\le (i_1+j_1)(i_2+j_2)\cdots(i_{1005}+j_{1005})$, where $i_1$ to $i_{1005}$ and $j_1$ to $j_{1005}$ are the integers from 1 to 2010. Then I proved that for any four positive numbers in increasing order, the smallest product of pairwise sums you can get is the product of the sum of the smallest two and the largest two. That means that if you add any n $i_n$ and $j_n$ that are not consecutive in the product, there will always be a smaller product by choosing $i_n$ and $j_n$ consecutive. The smallest product is thus the one where all the sums are sums of consecutive integers So I got the largest possible value as $(1+2)(3+4)(5+6)\cdots (2009+2010)=3\cdot 7\cdot 11\cdots 4019$.

Textangle wrote: I claimed that the lowest such upper bound is the maximum value of the product (I'm not certain about this though...) Unfortunately, claiming this without justification probably would not earn any points, since it's the critical part of your solution, and it seems like the main solution will not use this result anyways (so you won't get any points for "being on the right track").

andersonw wrote: Textangle wrote: I claimed that the lowest such upper bound is the maximum value of the product (I'm not certain about this though...) Unfortunately, claiming this without justification probably would not earn any points, since it's the critical part of your solution, and it seems like the main solution will not use this result anyways (so you won't get any points for "being on the right track"). Yeah, this is a HUGE part of the problem, and I would be very, very, very surprised if you got any more points for claiming that than for guessing the answer. (Which some people claimed would give a point IF you proved it was the upper bound)

I just set the inequalities as an equal sign. This guarantees (assuming a set of a_1, a_2... a_2010 fits it) that you get the maximum possible product by maximizing each a_n. Once you multiply together every a_i a_j = i + j, you take it to the 2009th root to obtain the product. To prove there is indeed a set of a_n's to satisfy this... I'm not sure how but I derived a method to obtaining them.

With your method, there are $\binom{2010}2$ equations and $2010$ variables, so it is very likely that not every equation is satisfied (and in fact, not every equation can be satisfied).

Yongyi781 wrote: With your method, there are $\binom{2010}2$ equations and $2010$ variables, so it is very likely that not every equation is satisfied (and in fact, not every equation can be satisfied). For a very straightforward proof that this cannot happen: a1a2 * a3a4 must equal a2a3*a1a4. If all the inequalities were equalities, then we would have 3 * 7 = 5*5.

dnkywin wrote: tjhance has a nice solution:

On another note, how many points do you think one would get if they constructed a sequence of the form $a_ia_{i+1}=2i+1$ AND used induction, but it was wrong? meh, didnt try induction. so i attempted to pull off a "CLEARLY THIS VALUE IS ACHIEVABLE AND NO EXPLANATION IS NEEDED" thing darn, could have gotten another problem but that is a nice solution

Textangle wrote: I got a solution for #3 ! Sketch- I saw that you could choose different ways to multiply 1005 inequalities together (so that each $a_i$ appears once in the left hand sides to get an upper bound for the product. I claimed that the lowest such upper bound is the maximum value of the product (I'm not certain about this though...) So, $a_1a_2a_3\cdots a_{2010}\le (i_1+j_1)(i_2+j_2)\cdots(i_{1005}+j_{1005})$, where $i_1$ to $i_{1005}$ and $j_1$ to $j_{1005}$ are the integers from 1 to 2010. Then I proved that for any four positive numbers in increasing order, the smallest product of pairwise sums you can get is the product of the sum of the smallest two and the largest two. That means that if you add any n $i_n$ and $j_n$ that are not consecutive in the product, there will always be a smaller product by choosing $i_n$ and $j_n$ consecutive. The smallest product is thus the one where all the sums are sums of consecutive integers So I got the largest possible value as $(1+2)(3+4)(5+6)\cdots (2009+2010)=3\cdot 7\cdot 11\cdots 4019$. I totally agree with this. And I got a solution that I think works: $ a_{1}= \sqrt{2+ \frac {1}{4}} $ $a_{2} = \frac{3} {\sqrt{2+ \frac {1}{4}}} $ $ a_{3}= \sqrt{6+ \frac {1}{12}}$ $a_{4} = \frac{7} { \sqrt{6+ \frac {1}{12}}} $ etc.

Hmm, according to Mathematica, a_1 = Limit[Sqrt[4*(n+1)-1]*Product[(4i-1)/(4i+1),1<=i<=n],n->infinity]=2*Gamma[5/4]/Gamma[3/4] works. Weird.

mathophile593 wrote: Hmm, according to Mathematica, a_1 = Limit[Sqrt[4*(n+1)-1]*Product[(4i-1)/(4i+1),1<=i<=n],n->infinity]=2*Gamma[5/4]/Gamma[3/4] works. Weird. That IS supposed to work. It agrees with my results, I think.....

Sorry, I think I'm missing something... $a_1a_{2010} \le 2011$ $a_2a_{2009} \le 2011$ $\dots $ $a_{1005}a_{1006} \le 2011$ So $a_1a_2...a_{2010} \le 2010^{1005}$

AK1024 wrote: Sorry, I think I'm missing something... $a_1a_{2010} \le 2011$ $a_2a_{2009} \le 2011$ $\dots $ $a_{1005}a_{1006} \le 2011$ So $a_1a_2...a_{2010} \le 2010^{1005}$ im not sure that achievable... especially since not only is, say, $a_1a_{2010} \le 2011$ but also $a_1a_{2009} \le 2010$, etc.

AK1024 wrote: Sorry, I think I'm missing something... $a_1a_{2010} \le 2011$ $a_2a_{2009} \le 2011$ $\dots $ $a_{1005}a_{1006} \le 2011$ So $a_1a_2...a_{2010} \le 2010^{1005}$ I think that is achievable, but the problem is that $3 * 7 * 11 * ... * 4019 \le 2010^{1005}$

IMO, peopel should differentiate between BOUNDS and MAXIMUMS/MINIMUMS. They are very different things

Textangle wrote: I got a solution for #3 ! Sketch- I saw that you could choose different ways to multiply 1005 inequalities together (so that each $a_i$ appears once in the left hand sides to get an upper bound for the product. I claimed that the lowest such upper bound is the maximum value of the product (I'm not certain about this though...) So, $a_1a_2a_3\cdots a_{2010}\le (i_1+j_1)(i_2+j_2)\cdots(i_{1005}+j_{1005})$, where $i_1$ to $i_{1005}$ and $j_1$ to $j_{1005}$ are the integers from 1 to 2010. Then I proved that for any four positive numbers in increasing order, the smallest product of pairwise sums you can get is the product of the sum of the smallest two and the largest two. That means that if you add any n $i_n$ and $j_n$ that are not consecutive in the product, there will always be a smaller product by choosing $i_n$ and $j_n$ consecutive. The smallest product is thus the one where all the sums are sums of consecutive integers So I got the largest possible value as $(1+2)(3+4)(5+6)\cdots (2009+2010)=3\cdot 7\cdot 11\cdots 4019$. And we have to explain why they can be reached.

What about generalization of this problem?

gold46 wrote: What about generalization of this problem? What generalizations did you have in mind? Generalizing to all even-length sequences is how one solves the problem. See whether you can use any of the methods in http://www.artofproblemsolving.com/Wiki/index.php/2010_USAMO_Problems/Problem_3 to obtain bounds for sequences with an odd length of 3 or more. (A sequence with just one term is not subject to any constraints). For three elements if: \begin{align*} xy &\le 3 \\ yz &\le 5 \\ zx &\le 4 \end{align*} We multiply all the inequalities to give $x^2y^2z^2 \le 60$, and so $xyz \le \sqrt{60}.$ Since in this case we can make all three inequalities be equalities, the maximum is $\sqrt{60},$ with $x = \sqrt{\frac{12}5}, y = \sqrt{\frac{15}4}, z = \sqrt{\frac{20}3}.$ For longer odd sequences it is probably not possible to make the product of the first and last term equal to $n+1,$ so the bound will be less than $\sqrt{(n+1)(2n-1)!!}.$ What is the bound for $n = 5$?

Phewww....it's too late to post but i have something, (however i didn't mention that a_{i}'s are all POSITIVE NUMBERS not POSITIVE INTEGERS,i worked with positive integers) the largest possible value of the product is $1^{1}*2^{2}*3^{4}*4^{4}*5^{6}*6^{6}*...*61^{62}*62^{62}*63^{27}$, and also I found it by substituting $a_{1},a_{2},...,a_{2010}$ are equal to $1,2,2,3,3,3,3,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,...,63,63,...,63$.which every positive integer number $i\leq 63$ appears $j$ times which $j$ here is the ammount of $i's$ exponent in the last answere i.e. $1^{1}*2^{2}*3^{4}*4^{4}*5^{6}*6^{6}*...*61^{62}*62^{62}*63^{27}$.

now i don't know how many points i get by this solution and also solve the problem in positive integers...

veyron wrote: I think $a_{2n-1}=2\sqrt{n}-\frac{1}{2\sqrt{n}} $, $a_{2n}=2\sqrt{n}$ for all $1\le n\le 1005$ also works I'd just like to point out that this is one of (many) solutions satisfying $a_i^2 \ge \frac{(2i-1)(2i+1)}{2i}$ for all $i$. Note that any such sequence satisfying $a_ia_{i+1}\le 2i+1$ for all $i$ automatically satisfies $a_ia_{i+j} \le 2i+j$ for $j>1$: just induct using the form $\frac{(a_ia_{i-j})(a_ia_{i+k})}{a_i^2}$---for simplicity, we can restrict ourselves to $k=j+1$ and $k=j$ (the intuition is that the inequalities $a_ia_j\le i+j$ are much more stringent when $|i-j|$ is small). After a little computation setting $a_i = c_i \sqrt{\frac{(2i-1)(2i+1)}{2i}}$ for $c_i\ge1$ (fix $k$ and consider $i=2k-1,2k,2k+1$, I believe we can actually WLOG assume $a_ia_{i+1} = 2i+1$ for all $i$ (but there are other ways to motivate that as well, like linear programming after taking logs), which makes things a bit simpler to think about.

Multiplying the all possible $\dbinom{2010}{2}$ combinations we get \[{\prod_{n=1}^{2010}\{a_n} < {(2011!/2!) (2012!/4!) (2013!/6!) ....(4019!/4018!)}^{1/2009}\]

As has been said before, that's a upper bound, not a maximum. You won't be able to achieve that.

On the https://www.artofproblemsolving.com/Wiki/index.php/2010_USAMO_Problems/Problem_3, should Wiki Solution wrote: $\begin{cases} a_ia_j=i+j=2i+1 & j=i+1, \\ a_ia_j = i+j = 2n-2 & j = i+2 = 2n, \\ a_ia_j < i+j &a... \end{cases}$ say $a_{2n-2}a_{2n}=4n-2$ (second line)?

My solution: I claim that the answer is $P = 3 \cdot 7 \cdots 4019$. We can see that the product cannot exceed this value because $a_1a_2 \le 3$, $a_3a_4 \le 7$ and so on up to $a_{2009}a_{2010} \le 4019$. Multiplying these gives us $P$ as a possible upper bound. Now I will show that this is achievable. We let $a_{2i}a_{2i-1} = 4i-1$ because we need that equality. Now let $a_{2i} = 2\sqrt{i}$, so that means we have $a_{2i-1} = 2\sqrt{i} - \frac{1}{2\sqrt{i}}$. Now we just need to check that these solutions do work. We need these four conditions to be true: $a_{2i}a_{2j} \le 2i + 2j \Rightarrow 4\sqrt{ij} \le 2i + 2j$ $a_{2i-1}a_{2j-1} \le 4i - 4j - 2 \Rightarrow 4\sqrt{ij} \ge \frac{(4i-1)(4j-1)}{4i-4j-2}$ $a_{2i-1}a_{2j} \le 2i+2j-1 \Rightarrow \frac{\sqrt{j}}{\sqrt{i}} \le \frac{2i+2j-1}{4i-1}$ $a_{2i}a_{2j-1} \le 2i+2j-1 \Rightarrow \frac{\sqrt{i}}{\sqrt{j}} \le \frac{2i+2j-1}{4j-1}$ The first one is easy AM-GM. Everything else can really be dealt with a little algebra. Just let $x = \sqrt{i}$ and $y = \sqrt{j}$. The second one: \begin{align*} 4xy (4x^2 + 4y^2 - 2) &\ge 16x^2y^2 - 4x^2 - 4y^2 + 1 \\ 16x^3y + 16xy^3 - 8xy &\ge 16x^2y^2 - 4x^2 - 4y^2 + 1 \\ 16xy(x^2 - xy + y^2) + 4(x - y)^2 &\ge 1 \end{align*} Notice that $x^2 - xy + y^2 \ge xy$ so we have the $LHS \ge 16x^2y^2 > 1$ for sure. The third one and fourth one are basically the same (symmetry). Again define $x, y$ like I did above. Then we have \begin{align*} \frac{y}{x} &\le \frac{2x^2 +2y^2 - 1}{4x^2 - 1} \\ 4x^2y - y &\le 2x^3 + 2xy^2 - x \\ x-y - 2x (x-y)^2 &\le 0 \\ (x-y)(1 - 2x^2 + 2xy) &\le 0 \end{align*} If $x = y$ we're all good. So suppose $x < y$. Then $2x^2 < 2xy$ so then $1 - 2x^2 + 2xy > 0$ for sure. Else $x > y$ and we need $2x^2 - 2xy - 1 \ge 0$. Now we need to go back to how we defined $x, y$. We thus have $2i - 2\sqrt{ij} - 1 \ge 0$. so we have $4i^2 - 4i + 1 \ge 4ij$. Which is the same as $i - 1 + \frac{1}{4i} \ge j$. But we know that $j \le i - 1$ so that inequality is indeed true.

My solution:

Obviously we can write \[a_1a_2 \cdots a_{2010} = (a_1a_2)(a_3a_4)\cdots (a_{2009}a_{2010}) \le (1+2)(3+4)\cdots (2009+2010).\]I claim that this bound is tight; i.e. there exist $a_i$ such that equality holds. Let's let $a_i^2 = b_i$, so $b_ib_j \le (i+j)^2$ for all distinct $i,j$. Let's make a preliminary choice of $b_i$ as follows: \[b_{2k+1} = 4k+2 + \frac{1}{4k+2}\]\[b_{2k+2} = 4k+4\] We have $b_{2k+1}b_{2k+2}> (4k+3)^2$. If we can show that $b_ib_j \le (i+j)^2$ for all $i < j$ not of the form $(i,j) = (2k+1, 2k+2)$, we can scale the $b_i$'s downwards until $b_{2k+1}b_{2k+2} = (4k+3)^2$ holds; then $b_ib_j \le (i+j)^2$ holds for all distinct $i,j$ and equality holds. To show that $b_ib_j \le (i+j)^2$ for all $i<j$ not of the form $(i,j) = (2k+1,2k+2)$, we will do casework on the parity of $i,j$. Case 1. $i,j$ are both even Proof. This is the easiest case. Just note that $b_i = 2i$ and $b_j = 2j$, so $b_ib_j = 4ij \le (i+j)^2$. Case 2. $i$ is even, $j$ is odd Proof. We have $b_i = 2i$ and $b_j = 2j + \frac{1}{2j}$, so $b_ib_j = 4ij + \frac{i}{j}$. Now we want $4ij + \frac{i}{j} \le i^2 + 2ij + j^2$, which rearranges to $(j-i)^2 \ge \frac{i}{j}$. But since $i<j$, we have $\frac{i}{j} < 1$. Furthermore, $j-i$ is an odd integer and thus $(j-i)^2 \ge 1$. It follows that $(j-i)^2 \ge \frac{i}{j}$. Caes 3. $i$ is odd, $j$ is even Proof. We have $b_i = 2i + \frac{1}{2i}$ and $b_j = 2j$, so $b_ib_j = 4ij + \frac{j}{i}$. Now we want $4ij + \frac{j}{i} \le i^2 + 2ij + j^2$, which rearranges to $(j-i)^2 \ge \frac{j}{i}$. If we decrease $i,j$ each by 2, $(j-i)^2$ stays constant but $\frac{j}{i}$ increases. Thus we can decrease $i$ until $i = 1$. We also know that $(i,j) \not= (2k+1, 2k+2)$, so $j-i \ge 3$. Thus $j \ge 4$. We want to prove $(j-1)^2 \ge \frac{j}{1} = j$, which is true for $j \ge 4$. Case 4. $i,j$ are both odd Proof. We have $b_i = 2i + \frac{1}{2i}$ and $b_j = 2j + \frac{1}{2j}$, so $b_ib_j = 4ij + \frac{j}{i} + \frac{i}{j} + \frac{1}{4ij}$. Now we want $4ij + \frac{j}{i} + \frac{i}{j} + \frac{1}{4ij} \le i^2 + 2ij + j^2$, which rearranges to $(j-i)^2 \ge \frac{j}{i} + \frac{i}{j} + \frac{1}{4ij}$. Similarly to above, we can decrease $i,j$ each by 2 until $i = 1$. Then we want to show that $(j-1)^2 \ge j + \frac{1}{j} + \frac{1}{4j}$. We know that $j \ge 3$, so $\frac{1}{j} + \frac{1}{4j} \le \frac{1}{3} + \frac{1}{12} = \frac{5}{12}$. We see that this is true for all $j \ge 3$, as desired. $\blacksquare$

Set $b_i=a_i^2/2, i=1,2,\dots,2010$. Then the restrictions are transformed as $$\displaystyle \sqrt{b_ib_j}\leq \frac{i+j}{2},\forall i\neq j\,\,\,\,\,\,\,\,\,\,(1)$$and we also search for the max possible value of $b_1b_2\cdots b_{2010}$. The idea is to match $b_i$'s by pairs $(b_{i_{k}}, b_{j_{k}} ),k=1,2,\dots,1005$ for which $(1)$ is equality. There is no guarantee it's possible, but we can try it, since if we succeed it would guarantee us the the product of $b_i$'s is maximal. Multiplying the inequalities $(1)$ for some pairing $ (b_{i_{k}}, b_{j_{k}} )$ we get $$\displaystyle \sqrt{b_1b_2\cdots b_{2010}}\le 2^{-1005}\prod_{k=1}^{1005}(i_k+j_k)\,\,\,\,\,\,\,\,\,\,(2)$$where $i_1,j_1,\dots, i_{1005},j_{1005} $ is some permutation of $1,2,\dots,2010$. This means the maximal value of $b_1b_2\cdots b_{2010}$ is no greater than the minimal value of the RHS of $2$ over all possible such permutations. In other words, we are doomed trying to group $b_i$'s into pairs $(b_{i_k},b_{j_k})$ for which $(1)$ becomes equality, unless $(i_k,j_k),k=1,2,\dots,1005$ minimizes the RHS of $(2)$. Playing a bit with it we suspect the minimum of the RHS of $(2)$ is attained for the pairing $(1,2),(3,4,),\dots,(2009,2010)$. There is no need to prove it, we only need to believe. Now, let us search for $b_1,b_2,\dots,b_{2010}$ satisfying $(1)$ and $$\displaystyle \sqrt{b_{2i+1}b_{2i+2}}=\frac{2i+1+2i+2}{2}=2i+3/2; i=0,1,\dots,1004$$ We proceed in a standard way, $b_{2i+2}=(2i+3/2)^2/b_{2i+1}$ and try to find appropriate values of $b_{2i+1},i=0,1,\dots,1004$ such that all inequalities in $(1)$ hold. For any 2 pairs $(2i+1,2i+2), (2j+1,2j+2), 0\le i<j\le 1004$ there are 4 possible combinations to check: \begin{align*} \displaystyle &\sqrt{b_{2i+1}b_{2j+1}}\leq i+j+1, \forall i<j\\ \displaystyle &\frac{(2i+3/2)(2j+3/2)}{\sqrt{b_{2i+1} b_{2j+1}} }\leq i+j+3, \forall i<j\\ \displaystyle &\frac{(2i+3/2)\sqrt{b_{2j+1}}}{\sqrt{b_{2i+1}}}\leq i+j+2, \forall i<j\\ \displaystyle &\frac{(2j+3/2)\sqrt{b_{2i+1}}}{\sqrt{b_{2j+1}}}\leq i+j+2, \forall i<j \end{align*}The first two inequalities yield $$\displaystyle \frac{(2i+3/2)(2j+3/2)}{i+j+3}\leq \sqrt{b_{2i+1}b_{2j+1}}\leq i+j+1, \forall i<j\,\,\,\,\,\,\,\,\,(3)$$The third and forth can be unified as $$\displaystyle \frac{(2i+3/2)\sqrt{b_{2j+1}}}{\sqrt{b_{2i+1}}}\leq i+j+2, \forall i\neq j\,\,\,\,\,\,\,\,\,(4)$$The upper bound of $(3)$ suggests $b_{2i+1}:=2i+1,i=0,1,\dots,1004$ would do the job because for those values the RHS of $(3)$ is just AM-GM inequality. Following this hunch, it remains to prove $(4)$ and the lower bound in $(3)$, which is straightforward, thou a bash. Stepping back, the corresponding maximal value of $a_1a_2\cdots a_{2010}$ is $\displaystyle \prod_{i=0}^{1024}(4i+3)$. $\blacksquare$ Comment. One could find more about motivation and other stuff in my blog here.

It is the obvious bound $3\cdot7\cdot11\cdots4019$, with upper bound obtained by \[a_1a_2\cdots a_{2010}=(a_1a_2)\cdots(a_{2019}a_{2020})\le3\cdots4019.\]Here are two constructions. In both, $a_{2n}a_{2n-1}=2n-1$, so equality holds. First, explicit construction. Let $a_{2n}=2\sqrt n$ and $a_{2n-1}=2\sqrt n-\frac1{2\sqrt n}$. We manually verify the required inequality: Let $i=2a$, $j=2b$, where $a<b$. Then \[a_ia_j=\left(2\sqrt a\right)\left(2\sqrt b\right)=4\sqrt{ab}<2(a+b).\] Let $i=2a-1$, $j=2b-1$, where $a<b$. Note the estimate \[\frac1{4\sqrt{ab}}\le\frac1{4\sqrt{a(a+1)}}<\frac2{\left(\sqrt{a+1}-\sqrt a\right)^2}=2\left(\sqrt{a+1}-\sqrt a\right)^2\le2\left(\sqrt b-\sqrt a\right)^2,\]From this, we have \[a_ia_j=\left(2\sqrt a-\frac1{2\sqrt a}\right)\left(2\sqrt b-\frac1{2\sqrt b}\right)<4\sqrt{ab}+\frac1{4\sqrt{ab}}-2\le2(a+b)-2.\] Let $i=2a-1$, $j=2b$, where $a\le b$. Then \[a_ia_j=\left(2\sqrt a-\frac1{2\sqrt a}\right)\left(2\sqrt b\right)=4\sqrt{ab}-\sqrt{\frac ba}\le2(a+b)-1.\] Let $i=2a$, $j=2b-1$, where $a<b$. Note the estimate \[\sqrt{\frac1b}=\frac2{2\sqrt b}\le\frac{2(b-a)}{\sqrt b+\sqrt a}=2\left(\sqrt b-\sqrt a\right)\implies 1-\sqrt{\frac ab}=2\left(\sqrt b-\sqrt a\right)^2.\]From this, we have \[a_ia_j=\left(2\sqrt a\right)\left(2\sqrt b-\frac1{2\sqrt b}\right)=4\sqrt{ab}-\sqrt{\frac ab}\le2(a+b)-1.\] Hence $a_ia_j\le i+j$ for all $i$, $j$. Second, more motivated construction. Take the sequence with $a_na_{n+1}=2n+1$ for each $n$ and $a_{2008}a_{2010}=4028$. Note that the inequality holds when $j-1=1$ by definition. We first verify the required inequality for $j-i=2$. Backwards induct on $i$, with the base case $i=2008$ given and the case $i=2007$ clear by noting \[a_{2007}a_{2009}=\frac{4015}{a_{2018}}\cdot\frac{4017}{a_{2010}}=\frac{4015\cdot4017}{4018}\le4017.\]If the inequality holds for integers greater than $i$, then \[a_ia_{i+2}=\frac{(a_ia_{i+1})(a_{i+2}a_{i+3})(a_{i+2}a_{i+4})}{(a_{i+1}a_{i+2})(a_{i+3}a_{i+4})}\le\frac{(2i+1)(2i+5)(2i+6)}{(2i+3)(2i+7)}\le2i+2,\]thus settling $j-i=2$. We now prove the inequality for all $i$, $j$, by induction on $j-i$ with increment $2$. The base cases $j-i\in\{1,2\}$ have already been proven. If $a_ia_j\le i+j$, then \[a_ia_{j+2}\le\frac{a_{j+2}}{a_j}(i+j)=\frac{a_{j+1}a_{j+2}}{a_ja_{j+1}}(i+j)=\frac{2j+3}{2j+1}(i+j)\le i+j+2,\]where the last inequality holds since \[\frac{2j+3}{2j+1}(i+j)\le i+j+2\iff\frac2{2j+1}(i+j)\le 2\iff i<j.\]This completes the proof.

I don't think the above proof has a valid proof of the bound for $a_ia_{i+2}$ (you can't cite $\le$ in the denominator). Here's a correct one? \[ a_{i}a_{i+2} = a_{i+2}a_{i+4} \cdot \frac{2i+1}{2i+3} \cdot \frac{2i+5}{2i+7} \le (2i+6) \cdot \frac{2i+1}{2i+3} \cdot \frac{2i+5}{2i+7} \le 2i + 2 \]

Easy problem Solution: a1×a2010<=2011 a2×a2009<=2011 . . . a1005×a1006<=2011 after multiplying our answer will be 2011¹⁰⁰⁵ I am shocked that l don't saw all solutions like this.