Take all indices $\mod 2n$. The only solution is $a_{2i-1}=1$ and $a_{2i}=2$ for $1\leq i \leq n$, which clearly works. The key is to sum the equations $a_{2k-1}=\frac{1}{a_{2k-2}}+\frac{1}{a_{2k}}$ (all sums from here will be from $1$ to $n$) \[\sum a_{2k-1}=\sum \frac{1}{a_{2k-2}}+\frac{1}{a_{2k}}=2\sum \frac{1}{a_{2k}}=2\sum \frac{1}{a_{2k-1}+a_{2k+1}}\]Now using the inequality $\frac{1}{a+b}\leq \frac{a+b}{4ab}$, \[\sum a_{2k-1}=2\sum \frac{1}{a_{2k-1}+a_{2k+1}}\leq \frac{1}{2}\sum \frac{1}{a_{2k-1}}+\frac{1}{a_{2k+1}}=\sum \frac{1}{a_{2k-1}}=\sum \frac{1}{\frac{1}{a_{2k-2}}+\frac{1}{a_{2k}}}\]Now using AM-HM, \[\leq \sum \frac{a_{2k-2}+a_{2k}}{4}=\frac{1}{2}\sum a_{2k}=\sum a_{2k-1}\]Since we have bounded above and below with $\sum a_{2k-1}$, the inqualities must be equalities! Looking at our C-S step, this implies that all the even indices are equal. This immediately gives all the odd indices are equal, so plugging in stuff we have $a_1=\frac{2}{a_2}$ and $a_2=2a_1$, which gives the solution.

tried for 2 hours, got nowhere.

We claim the only solution is $\boxed{(1,2,1,2,\cdots,1,2)}$ where $a_i = 1$ if $i$ is odd, $2$ if $i$ is even. Clearly this works. We note that $a_{2k+1} = \frac{1}{a_{2k}} + \frac{1}{a_{2k+2}} \Rightarrow \frac{2}{a_{2k+1}} = \frac{2}{\frac{1}{a_{2k}} + \frac{1}{a_{2k+2}}} \le \sqrt{a_{2k}a_{2k+2}} \Rightarrow \frac{a_{2k+1}}{2} \ge \frac{1}{\sqrt{a_{2k}a_{2k+2}}} \Rightarrow a_{2k+1} \ge \frac{2}{\sqrt{a_{2k}a_{2k+2}}}$ for all relevant $k$. $a_{2k} = a_{2k-1} + a_{2k+1} \ge 2\sqrt{a_{2k-1}a_{2k+1}} \ge 2\sqrt{\frac{2}{\sqrt{a_{2k-2}a_{2k}}} \cdot\frac{2}{\sqrt{a_{2k}a_{2k+2}}}} = \frac{4}{\sqrt{a_{2k}}\sqrt[4]{a_{2k-2}a_{2k+2}}} \Rightarrow a_{2k}^3 \sqrt{a_{2k-2}a_{2k+2}} \ge 16$. Taking product over all relevant $k$, we get $\left(\prod_{k=1}^n a_{2k}\right)^4 \ge 16^n$, thus $\prod_{k=1}^n a_{2k} \ge 2^n$. By AM-GM, we have $\sum_{k=1}^n a_{2k} \ge n \sqrt[n]{\prod_{k=1}^n a_{2k}} = 2n$. So average value of $a_{2k}$ is $\ge 2$. WLOG let $a_2 \ge 2$. Since $a_2 = a_1 + a_3 \ge 2$, WLOG let $a_3 \ge 1$. Since $a_3 = \frac{1}{a_2} + \frac{1}{a_4} \ge 1$ and $a_2 \ge 2$, then $a_4 \le 2$. Since $a_4 = a_3 + a_5 \le 2$ and $a_3 \ge 1$, then $a_5 \le 1$. Since $a_5 = \frac{1}{a_4} + \frac{1}{a_6} \le 1$ and $a_4 \le 2$, then $a_6 \ge 2$. Since $a_6 = a_5 + a_7 \ge 2$ and $a_5 \le 1$ then $a_7 \ge 1$. And so on. So we have $a_2 \ge 2, a_6 \ge 2, \cdots$, and we have $a_4 \le 2, a_8 \le 2, \cdots$. Notice if $n$ is odd, then at some point we get $a_2 \le 2$, $a_4 \ge 2$, and so on, so we get $a_2 = a_4 = \cdots = a_{2n} = 2$, which corresponds to the solution we mentioned at the beginning. If $n$ is even, then we have $a_2 + a_6 + \cdots + a_{2n-2} \ge 2\left(\frac{n}{2}\right) = n$. And we have $a_4 + a_8 + \cdots + a_{2n} \le 2\left(\frac{n}{2}\right) = n$. But notice that $a_2 + a_6 + \cdots + a_{2n-2} = a_4 + a_8 + \cdots + a_{2n} = a_1 + a_3 + \cdots + a_{2n-1}$. So $n \le a_1 + a_3 + \cdots + a_{2n-1} \le n, \Rightarrow a_1 + a_3 + \cdots + a_{2n-1} = n$. And so $a_2 + a_4 + \cdots + a_{2n} = (a_2 + a_6 + \cdots + a_{2n-2}) + (a_4 + a_8 + \cdots + a_{2n}) = n + n = 2n$. We have $2n = \sum_{k=1}^n a_{2k} = 2 \sum_{k=1}^n a_{2k-1} = 4\sum_{k=1}^n\frac{1}{a_{2k}}$. Let focus on this last term. We keep the sum $a_2 + a_4 + \cdots + a_{2n} = 2n$ constant, while reducing $\frac{1}{a_2} + \frac{1}{a_4} + \cdots + \frac{1}{a_{2n}}$ via smoothing. Notice that if $x,y \in \mathbb{R}^+$ and $x+y=\lambda$ is constant, then $\frac{1}{x} + \frac{1}{y} \ge \frac{2}{\frac{x+y}{2}} = \frac{4}{x+y} \iff \frac{x+y}{x} + \frac{x+y}{y} \ge 4 \iff \frac{x}{y} + \frac{y}{x} \ge 2$, true by AM-GM, equality holds iff $x = y$. Thus the whenever we have $a_{2i} \neq a_{2j}$ we can reduce the value of $\frac{1}{a_2} + \frac{1}{a_4} + \cdots + \frac{1}{a_{2n}}$. The minimum of this expression is then achieved when $a_2 = a_4 = \cdots = a_{2n} = 2$. And the minimum value of $4\left(\frac{1}{a_2} + \frac{1}{a_4} + \cdots + \frac{1}{a_{2n}}\right)$ is thus $4\left(\frac{n}{2}\right) = 2n$. But we need this to be the case, for the system of equations to be satisfied. So in fact system of equation can only be solved if $a_2 = a_4 = \cdots = a_{2n} = 2$. And thus if $n$ is even, the only solution is the one mentioned at the beginning as well.

I did something weird for this problem. I showed that if $a_{2k} > 2$, then either $a_{2k+2} < 2$ or $a_{2k-2}<2$, and if WLOG $a_{2k+2}<2$ then $a_{2k+4} >2$ and so on. This gives a contradiction iff n is odd (if $a_{2k} < 2$ its identical with the signs switched), so I think it's nontrivial, but couldn't generalize for $n$ even.

Guessed the equality case and managed to show the sum of the odds was at least $n$. Will I get a sympathy point?

Here's the (approximate) $\LaTeX$ for the equations: \begin{align*} a_{1} &=\frac{1}{a_{2 n}}+\frac{1}{a_{2}}, & a_{2}&=a_{1}+a_{3}, \\ a_{3}&=\frac{1}{a_{2}}+\frac{1}{a_{4}}, & a_{4}&=a_{3}+a_{5}, \\ a_{5}&=\frac{1}{a_{4}}+\frac{1}{a_{6}}, & a_{6}&=a_{5}+a_{7} \\ &\vdots & &\vdots \\ a_{2 n-1}&=\frac{1}{a_{2 n-2}}+\frac{1}{a_{2 n}}, & a_{2 n}&=a_{2 n-1}+a_{1} \end{align*}

The only solution is $a_{2k+1} = 1, a_{2k} = 2$, for $k$ integer. This works because \[1 = a_{2k+1} = \frac{1}{a_{2k}} + \frac{1}{a_{2k+2}} = \frac{1}{2} + \frac{1}{2} = 1\]\[2 = a_{2k} = a_{2k-1} + a_{2k+1} = 1 + 1 = 2\]Now we will show that these are the only solutions. Take the indices modulo $2n$ (so $a_{i} = a_{i+2n}$). I claim $a_{1} = a_{3}$. FTSOC, assume $a_{1}\neq a_{3}$, and WLOG let $a_{1} < a_{3}$. Now, we will prove by induction that \[(-1)^{k}\left[\frac{1}{a_{2k+4}} - \frac{1}{a_{2n-2k}}\right] = (k+1)(a_{3} - a_{1})+ k(a_{2n} - a_{4}) + (k-1)(a_{6} - a_{2n-2})\]\[+ (k-2)(a_{2n-4} - a_{8}) + (k-3)(a_{10} - a_{2n-6}) + \ldots + (-1)^{k-1}(a_{2n-2(k-1)} - a_{2k+2}) > 0\] For our base case of $k = 0$, we have \[\frac{1}{a_{4}} - \frac{1}{a_{2n}} = (a_{3} - a_{1}) > 0\]This is true by subtracting these two equations: \[a_{1} = \frac{1}{a_{2}} + \frac{1}{a_{2n}}, a_{3} = \frac{1}{a_{2}} + \frac{1}{a_{4}}\] Now onto our inductive step. Suppose that for all $i < k$, the above formula is true for $i$. Consider the following: \[a_{2k+2} = a_{2n-2k + 2} + (a_{2k+2} - a_{2n-2k+2})\]\[a_{2k+1} + a_{2k+3} = a_{2n-2k+1} + a_{2n-2k+3} + (a_{2k+2} - a_{2n-2k+2})\]\[\frac{1}{a_{2k}} + \frac{2}{a_{2k+2}} + \frac{1}{a_{2k+4}} = \frac{1}{a_{2n-2k}} + \frac{2}{a_{2n-2k+2}} + \frac{1}{a_{2n-2k+4}} + (a_{2k+2} - a_{2n-2k+2})\]If $k$ was even, then \[\frac{1}{a_{2k+4}} - \frac{1}{a_{2n}-a_{2k}} = 2\left[\frac{1}{a_{2n-2k+2}} - \frac{1}{a_{2k+2}}\right] + \frac{1}{a_{2n-2k+4}} - \frac{1}{a_{2k}} + (a_{2k+2} - a_{2n-2k+2})\]By our inductive hypothesis, we get that this is equal to \[2\left[k(a_{3} - a_{1}) + (k-1)(a_{2n} - a_{4}) + \ldots + (a_{2n-2k+4} - a_{2k})\right]\]\[- \left[(k-1)(a_{3} - a_{1}) + (k-2)(a_{2n} - a_{4}) + \ldots + (a_{2k-2} - a_{2n-2k+6})\right] + (a_{2k+2} - a_{2n-2k+2})\]\[= (k+1)(a_{3} - a_{1}) + k(a_{2n} - a_{4}) + \ldots + 2(a_{2n-2k+4} - a_{2k}) + (a_{2k+2} - a_{2n-2k+2})\]This completes the induction for $k$ even. We get a similar result for $k$ odd. This completes our induction. However, now we have to show that this result is greater than $0$. By our inductive hypothesis, if $i$ is odd, then $\frac{1}{a_{2n-2i}} - \frac{1}{a_{2i+4}} > 0 \Rightarrow a_{2i+4} > a_{2n-2i}$, so $a_{2i+4} - a_{2n-2i} > 0$. Similarly, for $i$ even, $a_{2n-2i} - a_{2i+4} > 0$. Since in the expression, $a_{3} - a_{1} > 0$, and for each $a_{2n-2i} - a_{2i+4}$ for $i$ even is greater than $0$, and for each $a_{2i+4} - a_{2n-2i}$ for $i$ odd is greater than $0$, we conclude this entire expression must be positive. This completes our induction. Now, since $(-1)^{k}(\frac{1}{a_{2k+4}} - \frac{1}{a_{2n-2k}}) > 0$, this means $a_{2k+4}\neq a_{2n-2k}$. If we find some $k,r$ such that $2n-2k = 2k+4 - 2nr$, this will yield a contradiction, since by taking modulo $2n$, this means $a_{2n-2k}, a_{2k+4}$ are the same variable, a contradiction. However, after simplifying, we get $n-k = k+2-nr$, or $n(r+1) = 2k+2$. Taking $k = n-1, r = 1$ gives the desired contradiction, so it is impossible to have $a_{3}\neq a_{1}$. Now, we have $a_{3} = a_{1}$. By a similar argument, we can show that $a_{1} = a_{3}; a_{3} = a_{5};\ldots a_{2n-1} = a_{1}$. Let $a_{2i+1} = T$. Then, $a_{2k} = a_{2k-1} + a_{2k+1} = 2T$, for all $k$. Now, \[T = a_{1} = \frac{1}{a_{2}} + \frac{1}{a_{2n}} = \frac{1}{2T} + \frac{1}{2T} = \frac{1}{T}\]Since $T > 0$, this means $T = 1$, so $a_{2k+1} = 1, a_{2k} = 2$ are our only solutions.

Note that $1 = \frac{a_{2i-1}}{a_{2i}} + \frac{a_{2i+1}}{a_{2i}}$. Thus, $n = \sum \frac{a_{2i-1}}{a_{2i}} + \frac{a_{2i+1}}{a_{2i}} = \sum \frac{a_{2i+1}}{a_{2i+2}} + \frac{a_{2i+1}}{a_{2i}} = \sum a_{2i+1}^2$. Also do this for the other side: $1 = \frac{1}{a_{2i+1}a_{2i}} + \frac{1}{a_{2i+1}a_{2i+2}}$, so $n = \sum \frac{1}{a_{2i+1}a_{2i}} + \frac{1}{a_{2i+1}a_{2i+2}} = \sum \frac{1}{a_{2i+1}a_{2i}} + \frac{1}{a_{2i-1}a_{2i}} = \sum \frac{1}{a_{2i+1}a_{2i-1}}$. Now use Holders: $\left ( \sum a_{2i+1}^2 \right )^{0.25} \left ( \sum a_{2i-1}^2 \right )^{0.25} \left ( \sum \frac{1}{a_{2i+1}a_{2i-1}} \right )^{0.5} \geq n$. But we have equality, so all $a_{\text{odd}}$ are equal, and this gives 1 2 1 2 ... as the only solution.

The only solution is when all odds are 1 and all evens are 2. Consider the odd $i$ such that $a_i$ has maximal distance from $1.$ If $a_i > 1,$ then one of the adjacent even values must be less than $2,$ let it be $a_{i-1}.$ Then, we can easily show that $|a_{i-2} - 1| > |a_i - 1|,$ contradiction. If $a_i < 1$, then one of the adjacent even values must be greater than $2,$ and we get a similar contradiction. Thus $a_i = 1$ and thus all odd indices are $1,$ so all evens are also $2$ and we're done.

Lemma 1: If $a<b$ and $c<d$ and $c=b+e$ and $d=a+f$, then $e<f$, for $a,b,c,d,e,f>0.$ Proof: $c<d$ so $b+e<a+f.$ Also, $a<b$ so $b-a>0.$ Then, $e<e+b-a=(b+e)-a<(a+f)-a=f.$ $\Box$ Lemma 2: If $m<n$ and $p<q$ and $p = \frac{1}{m} + \frac{1}{r}$ and $q = \frac{1}{n}+\frac{1}{s},$ then $s<r$ for $m,n,p,q,r,s>0.$ Proof: $m<n$ so $\frac{1}{m}>\frac{1}{n},$ so $\frac{1}{m}-\frac{1}{n} > 0.$ So $\frac{1}{r} < \frac{1}{r}+\frac{1}{m}-\frac{1}{n} < \frac{1}{n}+\frac{1}{s}-\frac{1}{n} = \frac{1}{s}.$ So $\frac{1}{r}<\frac{1}{s}$ and $s<r. \Box$ Now assume for contradiction that not all of $a_2, a_4, a_6, ..., a_{2n}$ are equal. Then there exist two neighboring terms that are not equal. WLOG let $a_2 > a_{2n}.$ Then, $a_1+a_3 > a_{2n-1}+a_1$ so $a_3 > a_{2n-1}.$ We apply Lemma 2 on $(m,n,p,q,r,s) = (a_{2n},a_2,a_{2n-1},a_{3},a_{2n-2},a_4)$ gives $a_4<a_{2n-2}.$ We apply Lemma 1 on $(a,b,c,d,e,f) = (a_{2n-1},a_3,a_{4},a_{2n-2},a_{5},a_{2n-3})$ gives $a_5 < a_{2n-3}.$ We continue alternately applying Lemma 2 and Lemma 1 until we eventually get $a_{n+1} > a_{n+1}.$ This is a contradiction, so our assumption was false. Therefore $a_2 = a_4 = ... = a_{2n}.$ Plugging this into the first $n$ equations gives $a_1 = a_3 = ... = a_{2n-1}.$ So $a_1 = \frac{2}{a_2}$ and $a_2 = 2a_1.$ Solving gives $a_1 = 1$ and $a_2 = 2.$ Therefore, our solution is $(1,2,1,2,...,1,2).$

Note that we have \[a_1^2=\frac{a_1}{a_{2n-1}+a_1}+\frac{a_1}{a_1+a_3}\] \[a_3^2=\frac{a_1}{a_{2n-1}+a_1}+\frac{a_1}{a_1+a_3}\]\[\cdots\] \[a_{2n-1}^2=\frac{a_{2n-1}}{a_{2n-3}+a_{2n-1}}+\frac{a_{2n-1}}{a_{2n-1}+a_1}\]and summing these gives \[a_1^2+a_3^2+\cdots+a_{2n-1}^2=n.\]We also have \[(a_1+a_3+\cdots+a_{2n-1})\cdot(a_1+a_3+\cdots+a_{2n-1}) \]\[=\left(\frac{2}{a_2}+\frac{2}{a_4}\cdots+\frac{2}{a_{2n}}\right)\left(\frac{a_2}{2}+\frac{a_4}{2}\cdots+\frac{a_{2n}}{2}\right)\]\[\geq n^2=(1+1+\cdots+1)(a_1^2+a_3^2+\cdots+a_{2n-1}^2).\]but this is the opposite of Cauchy Schwartz, so all the variables must be equal. It's easy to find the exact solutions from here.

yummy Answer is odd indexed terms 1, even indexed terms 2. Rewrite the problem in terms of the odd terms $b_k=a_{2k-1}$ which satisfy $$b_k=\frac{1}{b_k+b_{k-1}}+\frac{1}{b_k+b_{k-1}} \iff b_k^3+b_k^2(b_{k-1}+b_{k+1})+b_{k-1}b_kb_{k+1}=2b_k+b_{k-1}+b_{k+1}$$Observe by cauchy$$\left( \sum b_i \right)^2 = \left( \sum b_i+b_{i+1} \right) \left( \sum \frac{1}{b_i+b_{i+1}} \right) \ge n^2$$so the $b_i$ add to at least $n$. Main Claim: For any $k$ we have$$\text{min}(b_{k-1},b_{k+1}) \le \frac{2}{b_k}-b_k \le \text{max}(b_{k-1},b_{k+1})$$Proof: For convenience denote $b_k=t, b_{k-1}=x, b_{k+1}=y$ so by manipulating our relation we can get $$(x+t-\tfrac{1}{t})(y+t-\tfrac{1}{t})=\frac{1}{t^2}$$Considering $t$ as a constant this is a hyperbola in the $xy$ plane with vertices at $(-t,-t),(\tfrac{2}{t}-t,\tfrac{2}{t}-t)$. Since we only care about points in the first quadrant, we only look at the branch with the latter vertex. The claim is now clear. Also note that this means we must have $t< \sqrt{2}$, else both vertices of the hyperbola are in the third quadrant and there are no positive solutions. Now an extremal argument works: assume not all the $b_i$ are $1$, so their maxima is $b_k>1$; WLOG assume $b_{k+1}<b_{k-1}$ and for convenience denote $t=b_k, s=b_{k+1}$. So $s \le \frac{2}{t}-t$ and \begin{align*} \text{max}(b_k,b_{k+2}) &\ge \frac{2}{s}-s \ge \frac{2}{\frac{2}{t}-t}-\frac{2}{t}+t \\ &>t \end{align*}where we note $\frac{2}{x}-x$ is a decreasing function and the last inequality utilizes the fact that $\frac{2}{t}-t>0$.

Does this work? Substituting the equation in the second column into the equations in the first column, we have $$a_1=\frac{1}{a_{2n-1}+a_1}+\frac{1}{a_1+a_3},$$$$a_3=\frac{1}{a_1+a_3}+\frac{1}{a_3+a_5},$$$$\vdots$$$$a_{2n-1}=\frac{1}{a_{2n-3}+a_{2n-1}}+\frac{1}{a_{2n-1}+a_1}.$$Expanding the first two equations, we get $$a_1(a_{2n-1}+a_1)(a_1+a_3)=a_{2n-1}+2a_1+a_3,$$$$a_3(a_1+a_3)(a_3+a_5)=a_1+2a_3+a_5.$$Subtracting $a_{2n-1}+a_1$ from the first equation and $a_3+a_5$ from the second, we obtain $$a_1+a_3=(a_1^2+a_1a_3-1)(a_{2n-1}+a_1)=(a_3^2+a_1a_3-1)(a_3+a_5).$$Similarly, we can show that $$a_3+a_5=(a_3^2+a_3a_5-1)(a_1+a_3)=(a_5^2+a_3a_5-1)(a_5+a_7),$$$$\vdots$$$$a_{2n-1}+a_1=(a_{2n-1}^2+a_{2n-1}a_1-1)(a_{2n-3}+a_{2n-1})=(a_1^2+a_{2n-1}a_1-1)(a_1+a_3).$$For all $i,$ let $$x_i=a_{2i-1}^2+a_{2i-1}a_{2i+1}-1,$$$$y_i=a_{2i+1}^2+a_{2i-1}a_{2i+1}-1.$$Then, our equations become $$a_1+a_3=x_1(a_{2n-1}+a_1)=y_1(a_3+a_5),$$$$a_3+a_5=x_2(a_1+a_3)=y_2(a_5+a_7),$$$$\vdots$$$$a_{2n-1}+a_1=x_n(a_{2n-3}+a_{2n-1})=y_n(a_{1}+a_{3}).$$Now, note that \begin{align*} y_1(a_3+a_5) &= x_1(a_{2n-1}+a_1)\\ &= x_{1}x_{n}(a_{2n-3}+a_{2n-1})\\ &= x_{1}x_{n}x_{n-1}(a_{2n-5}+a_{2n-3})\\ &\vdots\\ &=x_1x_nx_{n-1}\dots x_{3}(a_3+a_5). \end{align*}Since $a_3,a_5>0,$ we know $a_3+a_5\ne 0.$ Therefore, $$y_1=\frac{x_{1}x_{2}\dots x_{n}}{x_2}.$$By multiplying all of our equations, we can show that $x_{1}x_{2}\dots x_{n}=1.$ Therefore, $y_{1}x_{2}=1,$ so by symmetry, $$x_{1}y_{2}=1,\hspace{50pt}y_{1}x_{2}=1,$$$$x_{2}y_{3}=1,\hspace{50pt}y_{2}x_{3}=1,$$$$\vdots\hspace{75pt}\vdots$$$$x_{n}y_{1}=1,\hspace{50pt}y_{n}x_{1}=1.$$This implies that $$x_1=x_3=\dots=\frac{1}{y_2}=\frac{1}{y_4}=\dots, $$$$x_2=x_4=\dots=\frac{1}{y_1}=\frac{1}{y_3}=\dots. $$If $n$ is odd, then all $x_i$ are equal and all $y_i$ are equal, so we must have $x_i=y_i$ for all $i.$ If $n$ is even, then $(x_1/y_1)^{n/2}=(x_{1}x_{2})^{n/2}=x_{1}x_{2}\dots x_{n}=1,$ so $x_1=\pm y_1$ and similarly for other $i.$ If $x_i=y_i$ for all $i,$ then $$a_{2i-1}^2+a_{2i-1}a_{2i+1}-1=a_{2i+1}^2+a_{2i-1}a_{2i+1}-1\implies a_{2i-1}=\pm a_{2i+1},$$from which it is easy to see that the only solution is $a_1=a_3=\dots=a_{2n-1}=1.$ If $x_i=-y_i$ for all $i,$ then $$a_{2i-1}^2+a_{2i-1}a_{2i+1}-1=-(a_{2i+1}^2+a_{2i-1}a_{2i+1}-1)\implies a_{2i-1}+a_{2i+1}=\sqrt{2},$$which yields no solutions.

I literally spent two hours on this and only managed to solve it for n=4

Huh, my solution seems to be different from everyone else's. Sketch: We claim by induction on $m \ge 2$ that $2^{-1/m} < a_{2i+1} < 2^{1/m}$ for all $i$. For base case, $a_{2i+1} = \frac{1}{a_{2i-1} + a_{2i+1}} + \frac{1}{a_{2i+1} + a_{2i+3}} < \frac{2}{a_{2i+1}}$ so $a_{2i+1} < 2^{1/2}$. Then $a_2i < 2*2^{1/2}$, then $a_{2i+1} = \frac{1}{a_{2i}} + \frac{1}{a_{2i+2}} > 2^{-1/2}$ For the inductive step we basically just do the same thing as the base case.

How much is proving for $n$ odd worth... I also did a trivial Cauchy bound.

@mc21s I used that same solution but distinguished between even and odd n. Beginning with $a_2>a_4$ yields $a_{n+3}>a_{n+3}$ eventually for odds, but I had to use the fact that $a_2+a_6+\cdots+a_{2n-2}=a_4+a_8+\cdots+a_{2n}$ (from the given equations) and $a_2+a_6+\cdots+a_{2n-2}>a_4+a_8+\cdots+a_{2n}$ (from the derived inequalities) to show that even n didn't work.

Am I the only one who proved $a_1 = 1$ with contradiction, which means all $a_i$ for odd $i$ is $1$? Everyone else did this cauchy stuff...

Take indices mod $2n$, and note that we have. Note that we have $$a_{2n}=a_{2n-1}+a_{2n+1}=\frac{2}{a_{2n}}+\frac{1}{a_{2n-2}}+\frac{1}{a_{2n+2}}$$Let $M,m$ denote the maximum and minimum values of $a_{\text{even}}$ respectively, so we have the inequalities $$M \leq \frac{2}{M}+\frac{2}{m} \leq m,$$so in fact we must have $M=m$, which also yields $M=m=2$. That is, $a_{2n}=2$ for all $n$. From this, we can easily get that $a_{2n+1}=\tfrac{1}{2}+\tfrac{1}{2}=1$ for all $n$ as well, hence the only solution is $(a_1,\ldots,a_{2n})=(1,2,\ldots,1,2)$, which clearly works. $\blacksquare$

Wasted 2 hours on spamming random inequality theorems to random expressions, bored, 1 hour break and after, immediate solve by another idea, duh Note that indices are taken $\mod 2n$. $\sum_{i=1}^n a_{2i-1}= 2\sum_{i=1}^n \frac{1}{a_{2i}}+ \frac{1}{a_{2i+2}} \stackrel{Titu}{\ge} \frac{2n^2}{\sum_{i=1}^n a_{2i}}$ Since $\sum_{i=1}^n a_{2i} = 2\sum_{i=1}^n a_{2i-1}$, we have $\sum_{i=1}^n a_{2i} \ge 2n$, $\sum_{i=1}^n a_{2i-1} \ge n$ I claim that $a_2=a_4=...=2,a_1=a+3=...=1$.Assume for contradiction that there is some $i$ such that $a_{2i}>2$. Then WLOG $a_{2i+1}>1$. Since $a_{2i+1} = \frac{1}{a_{2i}} + \frac{1}{a_{2i+2}}$ and $\frac{1}{a_{2i}} <\frac{1}{2}$, we must have $\frac{1}{a_{2i+2}}>\frac{1}{2} \Leftrightarrow a_{2i+2} <2$. We can similiarly prove $a_{2i+4}>2,a_{2i+6}<2$ etc. If $n$ is a odd number, repeating the prosses we will have $2>a_{2i}>2$, contraction, for even $n$, we have: $$ a_{2i}+a_{2i+4}....+ a_{2i-4} > a_{2i+2}+a_{2i+6}+...+a_{2i-2} \Leftrightarrow$$$$ (a_{2i-1}+a_{2i+1})+(a_{2i+3}+a_{2i+5})+...+ (a_{2i-5}+a_{2i-3})> (a_{2i+1}+a_{2i+3})+(a_{2i+5}+a_{2i+7})+...+ (a_{2i-3}+a_{2i-1})$$ But $LHS$ and $RHS$ are equal, contradicton. Hence we have $a_2=a_4=...=a_{2n}=2,a_1=a_3=...=a_{2n-1}=1$. $\ \blacksquare$

does assuming that $a_1<1$ ftsoc and wlogging $a_2\ge a_{2n}$ work? by inductively spamming the exact same logic over and over you get something like \[\cdots>a_7>a_3>a_1>a_5>a_9>\cdots\]and \[\cdots>a_{10}>a_6>a_2>a_4>a_8>\cdots\]but this seems really sketchy (if its right then by looping around u find everything must be equal which finishes) oops added detail you're supposed to repeat for $a_1>1$ to get that all odds are equal to $1$

asdf334 wrote: does assuming that $a_1<1$ ftsoc and wlogging $a_2\ge a_{2n}$ work? by inductively spamming the exact same logic over and over you get something like \[\cdots>a_7>a_3>a_1>a_5>a_9>\cdots\]and \[\cdots>a_{10}>a_6>a_2>a_4>a_8>\cdots\]but this seems really sketchy (if its right then by looping around u find everything must be equal which finishes) oops added detail you're supposed to repeat for $a_1>1$ to get that all odds are equal to $1$ How are you proving $a_3>a_1$? Also, the case where $a_1>1$ wouldn’t be the same I don’t think. If you add detail I can check your sol.

Consider the odd indexed term that is furthest away from 1. WLOG that this is $a_3$. If $a_3>1$, then we must have that one of $a_2$ and $a_4$ is less than 2, since their sum of reciprocals is greater than 1. WLOG that $a_2<2$. Then, $a_2=a_1+a_3,$ and $a_3>1$, so $a_1<1$. Additionally, $a_2$ is still less than 2, so $a_1$ is less than 1 by a larger margin than $a_3$ is greater than 1, which contradicts maximality of distance. We can do a similar thing in the other direction if $a_3<1.$ Therefore, the furthest odd indexed value from 1 is still equal to 1, so all odd values are 1, so all even values are 2. Therefore, the only solution is alternating 1 and 2, which clearly works. Remark: the motivation for this sol is when playing around with $n=4$, I found that if any odd-indexed value is not 1, we can derive other values that are even further away, even looping back, which cannot happen.

lrjr24 wrote: asdf334 wrote: does assuming that $a_1<1$ ftsoc and wlogging $a_2\ge a_{2n}$ work? by inductively spamming the exact same logic over and over you get something like \[\cdots>a_7>a_3>a_1>a_5>a_9>\cdots\]and \[\cdots>a_{10}>a_6>a_2>a_4>a_8>\cdots\]but this seems really sketchy (if its right then by looping around u find everything must be equal which finishes) oops added detail you're supposed to repeat for $a_1>1$ to get that all odds are equal to $1$ How are you proving $a_3>a_1$? Also, the case where $a_1>1$ wouldn’t be the same I don’t think. If you add detail I can check your sol. $a_3>a_1$ follows from $a_2>2a_1$ since we have $\frac{1}{a_2}\le \frac{a_1}{2}$ and $a_1<1$

All indices are modulo $2n$. Claim: For all $i$, $a_{2i} \le 2$. Proof: Suppose otherwise, so $a_{2x} > 2$ for some $x$. Then since $a_{2x} = a_{2x-1} + a_{2x+1}$, at least one of $a_{2x-1} > 1$ and $a_{2x+1} > 1$ must be true. Assume WLOG that $a_{2x+1} > 1$. Then from $a_{2x+1} = \frac{1}{a_{2x}} + \frac{1}{a_{2x+2}}$ and $\frac{1}{a_{2x}} < \frac12$, we must have $\frac{1}{a_{2x+2}} > \frac12$ giving $a_{2x+2} < 2$, or $a_{2x+1} + a_{2x+3} < 2$. Adding $1 < a_{2x+1}$ to this inequality gives $a_{2x+3} < 1$. Now $\frac{1}{a_{2x+2}} + \frac{1}{a_{2x+4}} < 1$, but $\frac{1}{a_{2x+2}} > \frac12$, giving $\frac{1}{a_{2x+4}} < \frac12$, or $a_{2x+4} > 2$, or $a_{2x+3} + a_{2x+5} > 2$. But $a_{2x+3} < 1$ so $a_{2x+5} > 1$ holds. At this point, the argument can be repeated from the beginning with $2x$ replaced by $2(x+2)$; doing so repeatedly gives that $a_{2x+k} > 2$ if $k \equiv 0 \pmod 4$, $a_{2x+k} > 1$ if $k \equiv 1 \pmod 4$, $a_{2x+k} < 2$ if $k \equiv 2 \pmod 4$, and $a_{2x+k} < 1$ if $k \equiv 3 \pmod 4$. If $n$ is odd, then $2n \equiv 2 \pmod 4$, so $2 < a_{2x} = a_{2x + 2n} < 2$, contradiction, so suppose $n$ is even. Then \[n < \sum_{i=1}^{n/2} a_{2x + 4i} = \sum_{i=1}^{n/2} a_{2x + 4i - 1} + a_{2x + 4i + 1} = \sum_{i=1}^n a_{2i-1} = \sum_{i=1}^{n/2} a_{2x + 4i + 1} + a_{2x + 4i + 3} = \sum_{i=1}^{n/2} a_{2x + 4i + 2} < n,\]contradiction. $\square$ The proof of the claim can be repeated with all inequality signs reversed to show that for all $i$, $a_{2i} \ge 2$, so $a_{2i} = 2$ for all $i$, giving $a_{2i-1} = 1$ for all $i$, which clearly works and is the only solution.

First, we have the following one-liner: \[ \sum_{k \ \text{odd}} a_k = 2 \left(\sum_{k \ \text{even}} \frac{1}{a_k} \right) = 2 \left(\sum_{k \ \text{odd}} \frac{1}{a_k + a_{k+2}}\right) \le \frac{1}{2} \left(\sum_{k \ \text{odd}} \frac{1}{a_k} + \frac{1}{a_{k+2}}\right) = \sum_{k \ \text{odd}} \frac{1}{a_k} = \sum_{k \ \text{even}} \frac{1}{\tfrac{1}{a_k} + \tfrac{1}{a_{k+2}}} \le \sum_{k \ \text{even}} \frac{a_k+a_{k+2}}{4} = \frac{1}{2} \sum_{k \ \text{even}} a_k. \]Realize that by summing the second column of equations, \[ \sum_{k \ \text{even}} a_k = 2 \sum_{k \ \text{odd}} a_k, \]and that our previous one-liner implies the \[ \sum_{k \ \text{even}} a_k \ge 2 \sum_{k \ \text{odd}} a_k. \]Thus all inequalities in the one-liner are tight, which implies $a_k=a_{k+2}$ for all even $k$. So, $(a_{2k})$ is a constant sequence (suppose that all terms are $a$) and $(a_{2k+1})$ is also a constant sequence (with all terms $2/a$). Thus $a = (2/a) + (2/a)$ so $a=2$. Hence, the only solution is $a_i = 2$ for even $i$ and $a_i=1$ for odd $i$, which clearly works.

probably fakesolved The only solution is $a_i=1$ for odd $i$ and $a_i=2$ for even $i.$ Otherwise summing all equations gives $\sum_{2\mid i}a_i=2\sum_{2\nmid i}a_i=4\sum_{2\mid i}\frac1{a_i}$ so $\sum_{2\mid i}a_i-\frac4{a_i}=0.$ Thus consider the maximum $a_i$ for even $i,$ call it $a_j.$ It must be $>2$ as otherwise we have $a_i=2$ for all even $i.$ Then letting $a_j=k$ and taking indices cyclically, we see one of $a_{j+1},a_{j-1}$ is $\ge \frac k2$ so assume without loss of generality that it is $a_{j+1}.$ Then we have $a_{j+2}\le\frac{2k}{k^2-2},a_{j+3}\le\frac{6k-k^3}{2k^2-4},a_{j+4}\ge\frac{k^3-2k}{-k^4+5k^2-2}$ if these are all positive. However, when $k > 2$ this is positive when $k \in \left(2,\sqrt{\frac{5+\sqrt{17}}2}\right)$ and we have $\frac{k^3-2k}{-k^4+5k^2-2}-k=\frac{k^3(k^2-4)}{-k^4+5k^2-2}>0$ contradicting maximality finishing.

Take indices mod $2n$ and note that \[a_{2k} = a_{2k-1}+a_{2k-3} = \frac{1}{a_{2k-2}}+\frac{2}{a_{2k}}+\frac{1}{a_{2k+2}}\]Let $m$ be the minimum of all $a_{2k}$ and $M$ be the maximum. We have \[m - \frac{2}{m} = \frac{1}{a_{2k-2}}+\frac{1}{a_{2k+2}} \ge \frac{2}{M}\]\[M-\frac{2}{M} = \frac{1}{a_{2\ell-2}}+\frac{1}{a_{2\ell+2}} \le \frac{2}{m}\]This implies $m \ge M$, so $m = M$. Therefore, all even indices of $a$ are the same. so all odd indices must also be the same. Now solving this system of two equations and variables gives $a_{2k} = 2$ and $a_{2k+1} = 1$. In other words, the only solution is to have all even index $a_i$ be $2$ and all odd index be $1$.

Here's my solution sketch (I'm posting here because I'm unsure whether this works). I have the full solution written out too if anyone wants it. Sketch: Claim: $a_n \le 1$ for all odd $n$. FTSOC, if, WLOG, $a_1 > 1$. We WLOG take $a_2 <= a_{2n}$ and get $a_2 \le \frac{2}{a_1}$. We then get $a_3 = a_2 + a_1 \le \frac{2}{a_1} - a_1 < \frac{1}{a_1}$. Continuing, we'd get $a_5 > a_1$, so if $n$ is even we'd induct to get $a_1 < a_5 < a_9 < \dots < a_{2n+1} = a_1$ (with indices taken mod 2n). However, if $n$ is odd, $a_1 > 1 \Longrightarrow a_3 < 1 \Longrightarrow a_5 > 1 \dots a_{2n+1} = a_1 < 1$ Claim: $a_n \ge 1$ for all odd $n$. If not then $a_{n+2} > 1$. (!) Either are impossible so $a_{odd} \equiv 1$ and $a_{even} \equiv 2$.

The answer is $(1, 2, 1, 2, \dots)$ only. To see this, reciprocating the second equation, we have $\frac 1{a_{2i}} = \frac 1{a_{2i-1} + a_{2i+1}}$, so \[a_1+a_3+\cdots+a_{2n-1} = 2\left(\frac 1{a_2}+\frac 1{a_4} + \cdots+\frac 1{a_{2n}}\right) = 2\left(\frac 1{a_1+a_3}+\frac 1{a_3+a_5}+\cdots+\frac 1{a_{2n-1}+a_1}\right).\]On the other hand, by reciprocating the first equation, we have \[\frac 2{a_{2i-1}} = \frac 2{\frac 1{a_{2i}}+\frac 1{a_{2i-2}}} \leq \frac{a_{2i}+a_{2i-2}}2.\]Summing these equations then yields \begin{align*} \frac 1{a_1}+\frac 1{a_3}+\cdots + \frac 1{a_{2n-1}} &\leq \frac 12\left(a_2+a_4+\cdots+a_{2n}\right) \\ &= a_1+a_3+\cdots+a_{2n-1} \\ &= 2\left(\frac 1{a_1+a_3}+\frac 1{a_3+a_5}+\cdots+\frac 1{a_{2n-1}+a_1}\right). \end{align*}By AM-HM we have $\frac 1{a_{2i-1}} + \frac 1{a_{2i+1}} \geq \frac 4{a_{2i-1}+a_{2i+1}}$, so equality must hold everywhere. It follows that $a_1=a_3=\cdots=a_{2n-1}$ and $a_2=a_4=\cdots=a_{2n}$ and the result follows.

USAMO 2021 p5. We claim that the only solution is $(1,2,1,2 \dots )$. Firstly we have that. $$a_{2k}=\frac{2}{a_{2k}} + \frac{1}{a_{2k-2}} + \frac{1}{a_{2k+2}}$$ Now let $a= max(a_{2k})$ and let $b=min(a_{2k})$. Then we have that. $$a=\frac{2}{a} + \frac{1}{a_{2k-2}} + \frac{1}{a_{2k+2}}$$And. $$b=\frac{2}{b} + \frac{1}{a_{2j-2}} + \frac{1}{a_{2j+2}}$$But we have that. $$a=\frac{2}{a} + \frac{1}{a_{2k-2}} + \frac{1}{a_{2k+2}} \leq \frac{2}{a} + \frac{2}{b}$$While we also have that. $$b=\frac{2}{b} + \frac{1}{a_{2j-2}} + \frac{1}{a_{2j+2}} \geq \frac{2}{b} + \frac{2}{a}$$Hence, we have that $b \geq a$ implying that the sequence $(a_{2i})$ is constant. Hence we are done.

Wow this was super nice The answer is $(a_{1}, a_{2}, \dots, a_{2n}) = (1, 2, 1, 2, 1, 2\cdots).$ Notice that summing all of the equations on the left gives that $$a_{1}+a_{3}+\dots+a_{2n-1} = \sum_{i=1}^{n}\frac{1}{a_{2i}} = \sum_{i=1}^{n}\frac{1}{a_{2i-1}+a_{2i+1}}.$$ Claim:For any positive reals $a,b,$ we have $$\frac{1}{a+b} \le \frac{a+b}{4ab},$$with equality if and only if $a=b.$ Proof:Cross multiplying, this is equivalent to having $4ab \le (a+b)^{2},$ or $(a-b)^{2}\ge0,$ which is obviously true. So, $$a_{1}+a_{3}+\dots+a_{2n-1} =\sum_{i=1}^{n}\frac{1}{a_{2i-1}+a_{2i+1}} \le \sum_{i=0}^{n} \frac{a_{2i+1}+a_{2i-1}}{4a_{2i-1}a_{2i+1}} = \frac{1}{2} \sum_{i=1}^{n} \frac{1}{a_{2i-1}}.$$Now, $$\sum_{i=1}^{n} a_{2i} = 2\sum_{i=1}^{n} a_{2i-1} \le 2\sum_{i=1}^{n} \frac{1}{a_{2i-1}} \le \sum_{i=1}^{n} \frac{2a_{2i}a_{2i+2}}{a_{i}+a_{i+2}} \le \sum_{i=1}^{n} \frac{a_{2i}+a_{2i+2}}{2} = \sum_{i=1}^{n}a_{2i}.$$This seems to be useless. However, it means that all of our inequalities must be equalities. So, $a_{1}=a_{3}=\cdots=a_{2n-1}=a$ and $a_{2}=a_{4} = \cdots = a_{2n}=b$ by the equality case of the claim. We get $b=2a,$ and $a=\frac{2}{b} = \frac{2}{2a}=\frac{1}{a},$ so $a=1,$ and $b=2.$ Hence, we get $a_{2i-1} = 1,$ and $a_{2i} = 2.$

Peculiar problem. I believe this is a new solution (that uses a more standard inequality approach.) We claim the only solution is $(a_1, a_2, \dots, a_{2n}) = (1, 2, 1, 2, \dots, 1, 2)$, which works by inspection. We will now prove it is the only one. Define $b_i = \dfrac{a_{2i}}{2}$, substituting the first group of equations into the second group yields $b_k = \dfrac{1}{b_k} + \dfrac{1}{2b_{k - 1}} + \dfrac{1}{2b_{k + 1}}$ for all $1 \le k \le n$ where indices are taken $\pmod{n}$. By Weighted AM-GM, we have \[2b_k = \dfrac{1}{b_k} + \dfrac{1}{2b_{k - 1}} + \dfrac{1}{2b_{k + 1}} \ge \dfrac{2}{\sqrt[4]{b_k^2 b_{k - 1} b_{k + 1}}} \implies b_k^6 b_{k - 1} b_{k + 1} \ge 1\]which multiplying cyclically gives $(b_1b_2 \cdots b_n)^8 \ge 1 \implies b_1b_2 \cdots b_n \ge 1$. Now summing the equations cyclically and dividing by $2$ yields \[b_1 + b_2 + \dots + b_n = \dfrac{1}{b_1} + \dfrac{1}{b_2} + \dots \dfrac{1}{b_n} \le \left( \dfrac{1}{b_1} + \dfrac{1}{b_2} + \dots \dfrac{1}{b_n} \right) (b_1 b_2 \dots b_n)^{2/n} = \sum_{\text{sym}} b_1^{2/n - 1} b_2^{2/n} \cdots b_n^{2/n}.\]But $(1, 0, \dots, 0) \succ \left( \dfrac{2}{n}, \dfrac{2}{n}, \dots, \dfrac{2}{n} - 1 \right)$ so by Muirhead the opposite of the above inequality must hold. Then equality occurs everywhere, thus $b_1 = b_2 = \dots = b_n \implies b_i = 1$ for all $i$. This gives the solution $(a_1, a_2, \dots, a_{2n}) = (1, 2, 1, 2, \dots, 1, 2)$ as claimed.