outline:

A friend of mine somehow used Jensen's. I'm not sure if it works; any idea on how it would run?

Induction: base case for $ n=2$ is simple. Let $ b$ be the minimal element. Now suppose $ a_1 > 4b$. Let $ (a_1+a_2...b)$ excluding $ a_n$ be $ r$, and let the sum of the reciprocals of those numbers be $ r'$. So we need to show that $ (r+a_n)(r'+1/a_n) > (n+\frac{3}{2})^2$ By induction, $ rr' > (n+\frac{1}{2})^2$. So we have to prove that $ r/a_n + a_nr' \geq 2n+1$. Now using the fact that $ x + 1/x \geq 2$, we may simplify this equation to: $ a_1/a_n + a_n/a_1 + b/a_n + a_n/b \geq 5$. We notice that since $ b$ is the smallest element, the sum is minimized by taking $ b$ as large as possible. But we assumed that $ b<a_1/4$. So take $ b=a_1/4$ and we need to show that $ a_1/a_n + a_n/a_1 + a_1/4a_n + 4a_n/a_1 \geq 5$. Now let $ a_1/a_n = q \geq 1$. Then $ q+1/q+q/4+4/q \geq 5 => 5q^2 - 20q + 20 \geq 0$ or $ q^2-4q+4 \geq 0$ which is true.

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JSteinhardt wrote: $ \frac {a}{1_n}$ One of the more humorous typos I've seen in awhile. And it's from Jacob Steinhardt too! First you let $ 1 = d$, and now you're putting subscripts on your $ 1$s. Next thing I know you'll be defining a function $ 1(x)$. Anyways, that's pretty nice, concise version of mine and some others I've seen, but you did skip over the proof of the $ n = 2$ case which you used at the end. Pretty big part of the problem, as that solution shows.

Well, that part was pretty trivial if you scaled both variables down, so that it was (x + 1)(1/x + 1) which is obviously increasing. Also, that was a pretty awesome proof .

my first reaction was to write out Lagrange's Identity, except I cited it as 'Legendre's identity' .. of course it was completely unnecessary, but I didn't feel like writing it over. Also looks very much like 2004 IMO Problems/Problem 4.

azjps wrote: Also looks very much like 2004 IMO Problems/Problem 4. lol yes that is the first thing I saw when I saw the problem

This is my solution verbatim. What do you think it will get? By the way, I left out the dollar signs around the equation and array environments. How do you make this work on AoPS?

This solution also uses nothing but AM-GM, which is interesting.

@JSteinhardt: that's exactly what i did except i managed to stretch it out to 2 pages instead of 2 lines, and my wording was poor and my grammar was bad. EDIT: If your solution is a little bit hard to follow, aka a not-so-great write-up, how many points will the graders dock? and if you mess up the wording of a clarification, how many points will the graders dock?

solved using induction. Suppose if a1 > 4a2 ,then the inequality doesn't hold for n = k-1 , then easy to prove it won't hold for n = k.

How about this solution:

blackbelt14253 wrote: @JSteinhardt: that's exactly what i did except i managed to stretch it out to 2 pages instead of 2 lines, and my wording was poor and my grammar was bad. EDIT: If your solution is a little bit hard to follow, aka a not-so-great write-up, how many points will the graders dock? and if you mess up the wording of a clarification, how many points will the graders dock? If it's still rigorous you should get full points. It just increases the danger that the graders can't understand your proof.

mathlord wrote: How about this solution: Thus, the maximum of the function occurs when each of the a's is either the maximum or the mininum. Hmm... this seems to be true if the function is convex, but did you find explicit equality cases? I know for a fact that 4 2 2 .... 2 2 1 is equality, and I'm pretty sure that is the only equality, which would contradict your statement.

Wlog, $\max(a_1, a_2, \ldots, a_n)=a_n$ and $\min(a_1, a_2, \ldots, a_n)=a_1$, hence we need $4 a_1\geq a_n$. Using Cauchy-Schwarz, $$ \left(n + \frac {1}{2}\right)^2 \geq \left(\sum_{i=1}^{n}a_i \right) \cdot \left(\sum_{i=1}^{n}\frac{1}{a_i} \right) \geq \left(\sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}}+n-2 \right)^2. $$Thus, we have $$\frac {5}{2}\geq \sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}}\Longleftrightarrow \frac{17}{4}a_1a_n\geq a_1^2+a_n^2\Longleftrightarrow \left(\frac{a_n}{4}-a_1\right)(4a_n-a_1)\leq 0$$Note that as $4a_n\geq a_n\geq a_1$, we must have $\dfrac{a_n}{4}-a_1 \leq 0\Longleftrightarrow 4a_1\geq a_n$. Done.

nice solution!

WLOG let $a_1\leq a_2\leq\dots \leq a_n$ and $a_1=1$. Now let $a_n>4$. We have $$(a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right)\geq \left(n-2+\sqrt{(1+a_n)\left(1+\frac{1}{a_n}\right)}\right)^2,$$to show a contradiction we only need to show that $\sqrt{(1+a_n)\left(1+\frac{1}{a_n}\right)}>\frac{5}{2}$ or $a_n+\frac{1}{a_n}>\frac{17}{4}=4+\frac{1}{4},$ which is trivial. $\blacksquare$

WLOG $a_1\le a_2\le\ldots\le a_n$. Assume FTSOC that $a_n>4a_1$. Denote $x=a_2+\ldots+a_{n-1}$, $y=\frac1{a_2}+\ldots+\frac1{a_{n-1}}$, and $z=\sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}}$. We have $z^2-2=\frac{a_1}{a_n}+\frac{a_n}{a_1}$ and $xy\ge(n-2)^2$ by AM-HM. After: \begin{align*}(a_1+a_n+x)\left(\frac1{a_1}+\frac1{a_n}+y\right)&=2+\frac{a_1}{a_n}+\frac{a_n}{a_1}+ya_1+ya_n+\frac x{a_1}+\frac x{a_n}+xy\\ &\ge2+z^2-2+2z\sqrt{xy}+xy\\ &=(z+n-2)^2\end{align*}it suffices to show that $z\ge\frac52$. But this is true since: $$z=\sqrt{\frac{a_1}{a_n}+\frac{a_n}{a_1}+2}=\sqrt{\frac{(a_n-4a_1)\left(a_n-4a_1+\frac{15}4a_1\right)}{a_1a_n}+\frac{25}4}\ge\sqrt{\frac{25}4}.~\square$$

Assume FTSOC that there exist indices $k$ and $\ell$ such that $\frac{a_k}{a_{\ell}} = d > 4$ and WLOG let $k=1$, $\ell=2$. Write $\left(\sum a_i\right)\left(\sum \frac{1}{a_i}\right)$ as $\frac{a_1}{a_2} + \frac{a_2}{a_1} + \sum_{i=3}^n\left( \frac{a_1}{a_i} + \frac{a_i}{a_1} + \frac{a_2}{a_i} + \frac{a_i}{a_2}\right) + \text{other stuff}$. We will deal with each term separately. First, $\text{other stuff}$ can be partitioned into pairs $\frac{a_i}{a_j} + \frac{a_j}{a_i}$, so the third term is at least $n^2 - 4n + 6$ by AM-GM. Next, by AM-GM $\frac{a_1}{a_i} + \frac{a_i}{a_2} > 2\sqrt{d}$ and $\frac{a_2}{a_i} + \frac{a_i}{a_1} > \frac{2}{\sqrt{d}}$. Since the function $x + \frac{1}{x}$ is increasing on the interval $(2,\infty )$ (just differentiate), the middle term is at least $5n - 10$. Finally, using the fact that $x + \frac{1}{x}$ is increasing on $(4,\infty )$, $\frac{a_1}{a_2} + \frac{a_2}{a_1} > \frac{17}{4}$. Putting it all together, $\left(\sum a_i\right)\left(\sum \frac{1}{a_i}\right) > \frac{17}{4} + 5n - 10 + n^2 - 4n + 6 = \left(n + \frac{1}{2}\right)^2$ and we have a contradiction, as desired.

WLOG $a_1\le a_2\le \cdots \le a_n$. Suppose FTSOC that $a_n>4a_1$. We have \[\left(\sum a_i\right)\left(\sum \frac{1}{a_i}\right)\ge \left(\sqrt{\left(a_1+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_n}\right)}+(n-2)\right)^2=\left( \sqrt{2+\frac{a_n}{a_1}+\frac{a_1}{a_n}}+(n-2)\right)^2 \]by Cauchy. Claim: $x+\frac{1}{x}\ge 4+\frac{1}{4}$ if $x\ge 4$. Proof: We need to show that in the interval $[4,\infty)$, $x+\frac{1}{x}$ is increasing. We have the derivative of $x+\frac{1}{x}$ is $1-\frac{1}{x^2}$. Since that is always greater than $0$ for $x\ge 4$, $x+\frac{1}{x}$ is increasing over $[4,\infty)$. $\blacksquare$ If we let $k=\frac{a_n}{a_1}$, we have $k>4$. Also, \begin{align*} \left(\sqrt{2+k+\frac{1}{k}}+(n-2)\right)^2 \\ >\left(\sqrt{2+4+\frac{1}{4}}+(n-2)\right)^2 \\ =\left(\sqrt{\frac{25}{4}}+(n-2)\right)^2 \\ =\left(\frac{5}{2}+(n-2)\right)^2 \\ =\left(n+\frac{1}{2}\right)^2 \\ \end{align*}a contradiction.

Assume WLOG $a_1 = \min(a_i)$ and $a_n = \max(a_i)$. From Cauchy, note that \[ \left(a_n + a_2 + \dots + a_{n-1} + a_1\right)\left(\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}\right)\ge \left(n-2 + \sqrt{\frac{a_n}{a_1}} + \sqrt{\frac{a_1}{a_n}}\right)^2.\]Thus, \[ \left(n +\frac{1}{2}\right)^2\ge \left(n-2 + \sqrt{\frac{a_n}{a_1}} + \sqrt{\frac{a_1}{a_n}}\right)^2\implies \frac{5}{2}\ge \sqrt{\frac{a_n}{a_1}} + \sqrt{\frac{a_1}{a_n}}.\]Substitute $t = \sqrt{a_n/a_1}\ge 1$. Now, define the function $f(t) = t + 1/t$ for $t > 0$. Note that $f$ is increasing in $[2,\infty)$, so \[ t + \frac{1}{t}\le \frac{5}{2}\implies f(t)\le f(2)\implies t\le 2,\]as desired.

Note $x+\frac{1}{x}$ is increasing over $(1,\infty)$. Define $Gain(a,b)$ for positive real numbers $a,b$ as $$Gain(a,b)=\frac{a}{b}+\frac{b}{a}-2.$$Clearly, $Gain(a,b)$ is always nonnegative. The condition is equivalent to $$\sum_{1\leq i<j\leq n}Gain(a_i,a_j)\leq n+\frac{1}{4}.$$For the rest of the solution, WLOG that $a_1=1$ is the smallest and that $a_n$ is the largest, since this is both symmetric and homogenous. Assume for the sake of contradiction that $a_n>4$. Then, clearly, $$Gain(a_1,a_n)> 4+\frac{1}{4}-2=\frac{9}{4}.$$ We then claim that for any $2\leq k\leq n-1$, we have $$Gain(a_1,a_k)+Gain(a_k,a_n)>1.$$This is just showing that $$\frac{1}{a_k}+a_k+\frac{a_k}{a_n}+\frac{a_n}{a_k}>5.$$Regroup this as $$a_k(1+\frac{1}{a_n})+\frac{1}{a_k}(1+a_n)>5.$$By AM-GM, $$a_k(1+\frac{1}{a_n})+\frac{1}{a_k}(1+a_n)\geq 2\sqrt{(1+\frac{1}{a_n})(1+a_n)}$$$$=2\sqrt{2+\frac{1}{a_n}+a_n}>2\sqrt{2+4+1/4}=5,$$which shows the claim. This means that $$Gain(a_1,a_n)+\sum_{k=2}^{n-1}Gain(a_1,a_k)+Gain(a_k,a_n)>\frac{9}{4}+(n-2)=n+\frac{1}{4}.$$This is only a subset of the total gain over all pairs, so the total gain is obviously larger than $n+\frac{1}{4}$ as well, contradiction, so we are done.

WLOG the $a_i$ are increasing. Observe the estimate $$(a_1+a_2+\cdots+a_n)\left(\frac 1{a_1} + \frac 1{a_2}+\dots + \frac 1{a_n}\right) \geq \left(n-2 + \sqrt{\frac{a_1}{a_n} + \frac{a_n}{a_1}}\right)^2.$$Set $k = \frac{a_n}{a_1}$ and analyzing the RHS in terms of $k$ yields $\frac 14 \leq \leq 4$, hence the result.

WLOG suppose $a_1$ is the maximum and $a_2$ is the minimum. Denote $x = \sqrt{\frac{a_2}{a_1}} + \sqrt{\frac{a_1}{a_2}}$. The problem statement reduces to proving $x \leq \frac 52$. The trick is to switch $a_1$ and $a_2$, then apply Cauchy to get \[\left(a_2 + a_1 + a_3 + \ldots + a_n\right) \left(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \ldots + \frac{1}{a_n}\right) \ge (n+x-2)^2.\] Hence, using our given bound, we get $x \leq \frac 52$.

FTSOC + WLOG, assume that $a_1=1$, and $a_2=4+\epsilon$ for some real number $\epsilon>0$. This ensures that $4\min(a_1,a_2,\dots,a_n)<\max(a_1,a_2,\dots,a_n)$. Then, we have the equation \[\left(1+[4+\epsilon]+a_3+a_4+\dots+a_n\right)\left(1+\frac{1}{4+\epsilon}+\frac{1}{a_3}+\frac{1}{a_4}+\dots+\frac{1}{a_n}\right)\]\[=\left([5+\epsilon]+a_3+a_4+\dots+a_n\right)\left(\left[\frac{5+\epsilon}{4+\epsilon}\right]+\frac{1}{a_3}+\frac{1}{a_4}+\dots+\frac{1}{a_n}\right),\]and by Cauchy-Schwarz; \[\left([5+\epsilon]+a_3+a_4+\dots+a_n\right)\left(\left[\frac{5+\epsilon}{4+\epsilon}\right]+\frac{1}{a_3}+\frac{1}{a_4}+\dots+\frac{1}{a_n}\right)\]\[\geq \left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+1+1+\dots+1\right)^2 = \left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+n-2\right)^2.\]However, since $\epsilon>0$, we have that \[4\epsilon^2+15\epsilon>0,\]\[\iff 100+40\epsilon+4\epsilon^2>25\epsilon+100,\]\[\iff 4(5+\epsilon)^2>25(4+\epsilon),\]\[\iff \frac{(5+\epsilon)^2}{4+\epsilon}>\frac{25}{4},\]\[\iff \sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}>\frac{5}{2},\]which implies that \[\left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+n-2\right)^2>\left(n+\frac{1}{2}\right)^2.\]Using this, we have that \[(a_1+a_2+\dots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\geq \left(\sqrt{\frac{(5+\epsilon)^2}{4+\epsilon}}+n-2\right)^2>\left(n+\frac{1}{2}\right)^2,\]a contradiction. Therefore, we must have $\max(a_1,a_2,\dots,a_n)\leq 4\min(a_1,a_2,\dots,a_n)$, finishing the problem.

We will prove the contrapositive; assuming that $\max > 4 \min$, we will show $\left( \sum a_i \right) \left( \sum \tfrac{1}{a_i} \right) > \left( n + \tfrac{1}{2} \right)^2$. WLOG $a_1 \leq a_2 \leq \dots \leq a_n$. Let $i$ be an index such that $a_1 \leq a_i \leq a_n$. Then, we have \begin{align*} & \left( \frac{a_1}{a_i} + \frac{a_i}{a_1} \right) + \left( \frac{a_n}{a_i} + \frac{a_i}{a_n} \right)\\ = ~ & \left( \frac{a_1}{a_i} + \frac{a_i}{a_n} \right) + \left( \frac{a_n}{a_i} + \frac{a_i}{a_1} \right) \\ \geq ~ & 2 \sqrt{ \tfrac{a_1}{a_n}} + 2 \sqrt{\tfrac{a_n}{a_1}} \\ > ~ & 2 \sqrt{ \tfrac{1}{4}} + 2 \sqrt{\tfrac{4}{1}} \\ = ~ & 5. \end{align*} Also, note that $\tfrac{a_1}{a_n} + \tfrac{a_n}{a_1}$ is strictly greater than $4 + \tfrac{1}{4} = \tfrac{17}{4}$. So, \begin{align*} & \left( \sum a_i \right) \left( \sum \tfrac{1}{a_i} \right) \\ = ~ & \sum_{1 \leq i \leq n} ( a_i \cdot \tfrac{1}{a_i} ) + \sum_{1 < i < j < n} \left( \tfrac{a_i}{a_j} + \tfrac{a_j}{a_i} \right) + \sum_{i \neq 1, n} \left( \tfrac{a_1}{a_i} + \tfrac{a_i}{a_1} + \tfrac{a_n}{a_i} + \tfrac{a_i}{a_n} \right) + \left( \tfrac{a_1}{a_n} + \tfrac{a_n}{a_1} \right) \\ > ~ & n + \binom{n-2}{2} \cdot 2 + 5(n-2) + \tfrac{17}{4} \\ > ~ & \left( n + \tfrac{1}{2} \right)^2. \end{align*}

Pretty cool inequality. Reminded me a bit of 24IMO4. We first show the case where $n=2$. Let $ka_1=a_2\ge a_1$. Then, \begin{align*} (a_1+a_2)\left(\frac{1}{a_1}+\frac{1}{a_2}\right) & \le \frac{25}{4}\\ k+\frac{1}{k} &\le \frac{17}{4}\\ 4k^2 - 17k+4 & \le 0\\ (4k-1)(k-4) & \le 0 \end{align*}Note that since $k\ge 1$, we must have $k\le 4$ so $a_2 \le 4a_1$ as desired. Now, for $n>2$ assume $a_1\le a_2 \le \dots \le a_n $ and $a_n > 4a_1$. Now, by Cauchy Schwarz and AM-GM we have, \begin{align*} (a_1+a_2+\dots + a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots + \frac{1}{a_n}\right) & = (a_2+\dots + a_{n-1})\left(\frac{1}{a_2}+\dots + \frac{1}{a_{n-1}}\right) + (a_1+a_n)\left(\frac{1}{a_2}+\dots + \frac{1}{a_{n-1}}\right) + (a_2+\dots + a_{n-1})\left(\frac{1}{a_1}+\frac{1}{a_n}\right) + (a_1+a_n)\left(\frac{1}{a_1}+\frac{1}{a_n}\right)\\ & \ge (n-2)^2 + 2\sqrt{ (a_1+a_n)\left(\frac{1}{a_1}+\frac{1}{a_n}\right)\left(\frac{1}{a_2}+\dots + \frac{1}{a_{n-1}}\right)(a_2+\dots + a_{n-1})} + (a_1+a_n)\left(\frac{1}{a_1}+\frac{1}{a_n}\right)\\ & \ge (n-2)^2 + 2\sqrt{ (a_1+a_n)\left(\frac{1}{a_1}+\frac{1}{a_n}\right)(n-2)^2} + (a_1+a_n)\left(\frac{1}{a_1}+\frac{1}{a_n}\right)\\ &> (n-2)^2 + 2\sqrt{\frac{25}{4}(n-2)^2} + \frac{25}{4}\\ &= n^2 - 4n + 4 +5(n-2) + 6 + \frac{1}{4}\\ & = n^2 + n + \frac{1}{4}\\ & = \left(n+\frac{1}{2}\right)^2 \end{align*}which is a clear contradiction. Thus if the desired inequality holds, we must have $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$, as desired.

Assume not. Then let $a_1$ be the smallest and $a_2$ be the largest. Then Cauchy on the rest gives $(n-2)^2$. Also, $\frac{a_k}{a_1}+\frac{a_2}{a_k}\ge 4$ and $\frac{a_k}{a_2}+\frac{a_1}{a_k}\ge 1$ by AM-GM. Between $a_1$ and $a_2$, their contribution is at least $2+4+\frac14$. Summing all of these gives exactly $(n+1/2)^2$.