inequality has returned, my prayers were not answered (i’m not in imo and prob trying later but it should be p1/4 difficulty)

Let's show that $a_{n+2}\geq a_n+3$ for all $n$. By C-S and AM-GM: $$a_{n+2}\geq a_n+\sqrt{\frac{(x_{n+1}+x_{n+2})^2}{x_{n+1}x_{n+2}}}>a_n+2.$$Therefore $a_{n+2}\geq a_n+3$. Since $a_1=1$, $a_{2023}\geq a_{2021}+3\geq\cdots\geq a_1+3033=3034$.

Motivation. It is obvious to see that $3034=2022\times\frac 32+1,$ which leads us to the lemma below$.$ Lemma. For $\forall n\in\mathbb Z_+,a_{n+2}\geq a_n+3.$ Lemma Proof. As $a_n,a_{n+2}\in\mathbb Z_+,$ we only need to prove that $a_{n+2}>a_n+2.$ Using Cauchy inequality$,$ $$\begin{aligned}a_{n+2}=\sqrt{\sum\limits_{k=1}^{n+2}x_k\sum\limits_{k=1}^{n+2}\frac 1{x_k}}&\geqslant\sqrt{\sum\limits_{k=1}^{n}x_k\sum\limits_{k=1}^{n}\frac 1{x_k}}+\sqrt{(x_{n+1}+x_{n+2})\left(\frac 1{x_{n+1}}+\frac 1{x_{n+2}}\right)}\\&=a_n+\sqrt{4+\frac{(x_{n+1}-x_{n+2})^2}{x_{n+1}x_{n+2}}}>a_n+2.\blacksquare\\\end{aligned}$$Proof. Using Lemma we have $a_{2023}\geq a_{2021}+3\geq\cdots\geq a_1+3\times 1011=3034.\blacksquare$

We claim $a_{n + 2} \ge a_n + 3$ which is clearly enough. Let $s_n = x_1 + x_2 + \dots + x_n$ and $h_n = \frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}$. Then \[a_{n + 1}^2 = s_{n + 1}h_{n + 1} = (s_n + x_{n + 1})\left(h_n + \frac{1}{x_{n + 1}}\right) = s_nh_n + \left(\frac{s_n}{x_{n + 1}} + x_nh_{n + 1}\right) + 1 = a_n^2 + \left(\frac{s_n}{x_{n + 1}} + x_{n + 1}h_n\right) + 1\] By AM-GM we have $\frac{s_n}{x_{n+1}} + x_{n+1}h_n \ge 2\sqrt{s_nh_n} = 2a_1$ and therefore $a_{n + 1}^2 \ge a_n^2 + 2a_n + 1 = (a_n + 1)^2$. Thus $a_{n + 1} \ge a_n + 1$ for all $n$, with equality if and only if $\frac{s_n}{x_{n+1}} = x_{n+1}h_n$, or $x_{n+1}^2 = \frac{s_n}{h_n}$. We finally prove that equality cannot happen twice in a row. Otherwise we have \[x_{n + 1}^2 = \frac{s_{n + 1}}{h_{n + 1}} = \frac{s_n + x_{n + 1}}{h_n + \frac{1}{x_{n + 1}}} = \frac{s_n + \sqrt{\frac{s_n}{h_n}}}{h_n + \sqrt{\frac{h_n}{s_n}}} = \frac{s_n}{h_n} = x_n^2\] A contradiction as $x_1, x_2, \dots, x_{2023}$ are all distinct.

Solution : We will move on with a claim first. Claim : $a_{n+2} > a_n+2$ Proof : Observe that $$a_{n+2} > a_n+2$$$$ \Longleftrightarrow \sqrt{(x_1+....+x_{n+2})(\frac{1}{x_1}+....+\frac{1}{x_{n+2}}} > \sqrt{(x_1+....+x_{n})(\frac{1}{x_1}+....+\frac{1}{x_{n}}} +2 .$$Now let $y=\sum_{i=1}^{n} x_i$ and $z=\sum_{i=1}^{n} \frac{1}{x_i}$ , then we have to show $$\sqrt{(y+x_{n+1}+x_{n+2})(z+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} > \sqrt{yz}+2$$$$ \Longleftrightarrow yz+y(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})+z(x_{n+1}+x_{n+2})+(x_{n+1}+x_{n+2})(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}) > yz+4\sqrt{yz}+4$$and it is obvious by AM-GM. Hence we get $$a_{2023} \geq a_{2021}+3 \geq a_{2019}+6 \geq \cdots a_4+3030 \geq a_1+3033 =1+3033=3034 .$$Note : The problem was very good but the main part was to prove $a_{n+2} \geq a_{n}+3$ using AM-GM and Algebraic Expansion.

Obviously, $a_n$ should be an increasing (strictly) sequence of positive integers with $a_1=1$. We will show that $a_{n+1}=a_n+1$ and $a_{n+2} = a_{n+1}+1$ doesn't hold at the same time. Assume it does. We know that \[ a_{n+1}^2-a_n^2=x_{n+1}\left(\frac{1}{x_1}+\ldots +\frac{1}{x_{n}}\right) + \frac{1}{x_{n+1}} \left(x_1+\ldots +x_n\right)+1 = 2a_n+1 \]and \[ 2a_n = x_{n+1}\left(\frac{1}{x_1}+\ldots +\frac{1}{x_{n}}\right) + \frac{1}{x_{n+1}} \left(x_1+\ldots +x_n\right) \ge 2a_n \]by AM-GM, so we should have \[ x_{n+1}\left(\frac{1}{x_1}+\ldots +\frac{1}{x_{n}}\right) = \frac{1}{x_{n+1}} \left(x_1+\ldots +x_n\right) = a_n. \]Similarly, we can obtain \[ x_{n+2}\left(\frac{1}{x_1}+\ldots +\frac{1}{x_{n+1}}\right) = \frac{1}{x_{n+2}} \left(x_1+\ldots +x_{n+1}\right) = a_{n+1}. \]That means \[ a_nx_{n+1}+x_{n+1}=x_1+x_2+\ldots +x_n+x_{n+1} = a_{n+1}x_{n+2} \]and since $a_{n+1}=a_n+1$ we get that $x_{n+1}=x_{n+2}$, which is a contradiction. Hence, if $a_{n+1}-a_n=1$ then $a_{n+2}-a_{n+1}>1$, or vice versa. In any case, $a_{n+2}-a_n\ge 3$ and $a_{2023} \ge a_1+3\cdot 1011 = 3034$.

Solution without using AM-GM Lets prove $a_{n+1}$$\geq$$a_{n-1}+3$ s = $x_1 + x_2 + ... + x_{n-1}$ h = $\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_{n-1}}$ $a^2_{n-1}$ = hs $a^2_{n} = (s+x_n)(h+\frac{1}{x_n})$ $a^2_{n} = hs + \frac{s}{x_n} + {x_n}h + 1$ $a^2_{n-1} = a_{n-1}^2 + \frac{s}{x_n} + xh + 1$ assume $a^2_{n}$=$a_{n-1}+1$;$a^2_{n+1}$=$a_{n}+1$ otherwise its trivial since obviously $a_{n-1}<a_{n}$ 2$a_{n-1}=\frac{s}{x_n}+xh$ $h=\frac{a^2}{s}$ 2$a_{n-1}=\frac{s}x+\frac{x}s{a_{n-1}^2}$ let $k = \frac{s}x$ 2$a_{n-1}$=k+$\frac{1}k{a_{n-1}^2}$ solving for $a_{n-1}$, we get: $a_{n-1}$=$\frac{s}x$ s=$a_{n-1}$x $a_{n-1}+1=\frac{s+x}x'$ x'(a+1)=x(a+1) x'=x contradiction, so $a_{n+1}$$\geq$$a_{n-1}+3$

The crucial claim is the following: Claim: For any $n$, it cannot be the case that $a_{n+1} = a_n + 1$ and $a_{n+2} = a_{n+1}+1$. Proof: Suppose, FTSOC, $a_{n+1} = a_n + 1, a_{n+2} = a_{n+1}+1$. $$ a_{n+1} = a_n + 1 \iff x_{n+1} \left( x_1 + x_2 + \cdots + x_n \right) + \dfrac{1}{x_{n+1}} \left(x_1 + \cdots + x_n \right) = 2a_n$$ Treat this as a quadratic in $x_{n+1}$: its discriminant vanishes and the only root is $x_{n+1} = \dfrac{a_n}{1/x_1 + \cdots + 1/x_n}$. Similarly, $x_{n+2} = \dfrac{a_{n}+1}{1/x_1 + \cdots + 1/x_{n+1}}$. I now claim that $x_{n+1} = x_{n+2}$, which is a contradiction. This follows from a simple calculation: $$ x_{n+1} = x_{n+2} \iff a_n \left( \dfrac{1}{x_1} + \cdots + \dfrac{1}{x_{n+1}}\right) = (a_n+1) \left( \dfrac{1}{x_1} + \cdots + \dfrac{1}{x_n} \right) \iff x_{n+1} = \dfrac{a_n}{1/x_1+\cdots + 1/x_n} \blacksquare$$ Now, note that $a_2 \geq 2$ by Cauchy-Schwartz. Since equality cannot hold (as $x_1, x_2$ are distinct), in fact, $a_2 \geq 3$. Also note that $a_n$ is strictly increasing. The claim shows that $a_i$ must jump more than one once every two steps, i.e., $a_{n+2} \geq a_n + 3$. Applying this inequality repeatedly gives the result.

$$a_{n+1}^2 = a_n^2 + 1 + x_{n+1}(\frac{1}{x_1}+ \cdots + \frac{1}{x_{n}}) + \frac{1}{x_{n+1}}(x_1 + \cdots + x_{n}) \geq a_n^2 + 1 + 2a_n = (a_n+1)^2$$By AM-GM, with equality if and only if the two things we are summing are equal, in particular $x_{n+1} = \sqrt{\frac{A}{B}}$ where $A = \sum a_i$ and $B = \sum \frac{1}{a_i}$. If also $a_{n+2} = a_{n+1} + 1$ then again we must have $x_{n+2} = \sqrt{\frac{A+\sqrt{\frac{A}{B}}}{B+\sqrt{\frac{B}{A}}}}$ but both things inside the square root are equal so $x_{n+1} = x_{n+2}$ but that cant happen so at least $a_{n+2} \geq a_n + 3$. Finally $a_{2023} \geq 1 + 3 \cdot 1011 = 3034$ as we wanted to show.

Sneaky little bugger. Define $A_n := x_1 + x_2 + \cdots + x_n$ and $B_n := \tfrac 1{x_1} + \tfrac1{x_2} + \cdots + \tfrac1{x_n}$. Observe that \begin{align*} a_{n+1}^2 &= \left(A_n + x_{n+1}\right)\left(B_n + \frac{1}{x_{n+1}}\right) \\ &= A_nB_n + \frac{A_n}{x_{n+1}} + B_n x_{n+1} + 1 \\ &\geq A_nB_n + 2\sqrt{A_nB_n} + 1 = (a_n + 1)^2. \end{align*}Hence $a_{n+1} \geq a_n + 1$. We now prove the stronger inequality $a_{n+2} \geq a_n + 3$. Suppose, by way of contradiction, that $a_{n+2} = a_n + 2$. Then equality cases in both inequalities above (i.e. for $a_{n+1}$ and $a_{n+2}$) hold, so $x_{n+1} = \sqrt{\tfrac{A_n}{B_n}}$ and \[ x_{n+2}^2 =\frac{A_n + x_{n+1}}{B_n + \frac 1{x_{n+1}}} = \frac{A_n + \sqrt{\frac{A_n}{B_n}}}{B_n + \sqrt{\frac{B_n}{A_n}}}= \frac{A_n}{B_n}. \](The last equality is true because $\tfrac{x_{n+1}}{\frac{1}{x_{n+1}}} = x_{n+1}^2 = \tfrac{A_n}{B_n}$.) This contradicts the fact that the $x_j$ are pairwise distinct. Remark. I think the claims here are motivated, but the main sticking point for me was getting over my confusion about why the bound wasn't tighter. (It didn't help that I made a few algebra mistakes along the way.) Proving the inequality $a_{n+2} \geq a_n + 3$ directly is certainly faster, but this solution shows why we need to consider this stronger bound in the first place.

MOTIVATION : The thing which motivate $a_{n+2} \geq a_{n}+3$ is $a_{2023}\geq 3034$ as actually this is the point for me which makes me realise to follow up this claim. proving that claim is same as GrantStar did and finishes up the problem.

So. Is the bound $3034$ actually sharp?

Too easy for IMO inequality. Claim : $a_{n+2}>a_{n}+2$ which is equivalent to $a_{n+2} \ge a_{n}+3$ since $a_n$ is an integer. Hence $$a_{2023} \ge a_1+\sum_{n=1}^{1011}3=3034$$ Then prove the claim: We define $X_n=x_1 + x_2 + \cdots + x_n$ and $Y_n=\frac{1}{x_1}+\cdots+\frac{1}{x_{n}}$, so $a_n=\sqrt{X_n Y_n}$ \begin{align*} a_{n+2}=&\sqrt{(x_1 + x_2 + \cdots + x_n+x_{n+1}+x_{n+2}) (\frac{1}{x_1}+\cdots+\frac{1}{x_{n}}+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} \\ &= \sqrt{(X_n+x_{n+1}+x_{n+2})(Y_n+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} \\ &= \sqrt{a_n^2+X_n(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})+Y_n(x_{n+1}+x_{n+2})+(x_{n+1}+x_{n+2})(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} \\ &> \sqrt{a_n^2+ 4 \sqrt{X_nY_n}+4} \\ &= \sqrt{a_n^2+4a_n+4}=a_n+2 \end{align*}

Let $S_n=\sum_{i=1}^{n}x_i$ and $H_n=\sum_{i=1}^{n}1/x_i$. Then we have that: $a_{n+2}^2$ $=$ $S_{n+2}H_{n+2}$ $=(S_n+(x_{n+1}+x_{n+2}))(H_n+(1/x_{n+1}+1/x_{n+2}))$ $\geq$ $[\sqrt{S_nH_n}+\sqrt{(x_{n+1}+x_{n+2})(1/x_{n+1}+1/x_{n+2})}]^2$ $>$ $(\sqrt{S_nH_n}+2)^2$ $=(a_n+2)^2$, hence $a_{n+2}>a_n+2$. Now using that $a_n$ is always integer we have that $a_{n+2}$ $\geq$ $a_n+3$ and inductively we get that $a_{2023} \geq a_1+3*1011=3034$, as needed. Note that both the 1st and the 2nd inequality I used hold from C-S and the equality cannot hold because $x_i's$ are different

GrantStar wrote: Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that \[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$ Isn't this equivalent to "if (...impossible condition...) then (...anything at all...)"? In particular, when $n=2$, you've got $a_2^2=(x_1+x_2)\left(\frac{1}{x_1}+\frac{1}{x_2}\right)=\frac{(x_1+x_2)^2}{x_1x_2}$, which is never an integer for positive integers $x_1\ne x_2$.

oolite wrote: GrantStar wrote: Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that \[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$ Isn't this equivalent to "if (...impossible condition...) then (...anything at all...)"? In particular, when $n=2$, you've got $a_2^2=(x_1+x_2)\left(\frac{1}{x_1}+\frac{1}{x_2}\right)=\frac{(x_1+x_2)^2}{x_1x_2}$, which is never an integer for positive integers $x_1\ne x_2$. Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive REAL numbers

oolite wrote: GrantStar wrote: Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that \[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$ Isn't this equivalent to "if (...impossible condition...) then (...anything at all...)"? In particular, when $n=2$, you've got $a_2^2=(x_1+x_2)\left(\frac{1}{x_1}+\frac{1}{x_2}\right)=\frac{(x_1+x_2)^2}{x_1x_2}$, which is never an integer for positive integers $x_1\ne x_2$. Why are $x_1$ and $x_2$ integers?

RevolveWithMe101 wrote: Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive REAL numbers Thank you! I couldn't quite believe what I thought I was seeing! Sanity has returned.

Notice $a_1=1$. We claim that $a_{n+2}-a_n > 2$, which would imply $a_{n+2}-a_n \ge 3$ and give $a_{2023} \ge 2 \cdot 1011 + 1$, as desired. Using Cauchy, we find \begin{align*} a_{n+2} &\ge \sqrt{a_n^2+4+\left(x_1+x_2\right)\left(\frac{1}{x_3}+\ldots+\frac{1}{x_n}\right) + \left(x_3+\ldots+x_n\right) \left(\frac{1}{x_1}+\frac{1}{x_2}\right)} \\ &\ge \sqrt{a_n^2+4+4a_n} = a_n+2, \end{align*} but equality only holds when $x_1 = x_2$, which we cannot have. $\blacksquare$

Let $p=x_1+x_2+...+x_n$ and $q=\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}$ $a_{n+2}=\sqrt{(p+x_{n+1}+x_{n+2})(q+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} \ge \sqrt{(\sqrt{pq}+1+1)^2} = a_n+2$, but this is only equal when the quotient between the terms are the same. The values of the quotients are $x_{n+1}^2$ and $x_{n+2}^2$, which are different. So $a_{n+2} \ge a_n+3$ Then because $a_1=1$ then repeatedly use this to get that $a_{2023} \ge 3034$

Claim 1. $a_2 \geq 3$. Proof. We have \begin{align*} a_2 &= \sqrt{(x_1+x_2) \left(\dfrac{1}{x_1} + \dfrac{1}{x_2}\right)} \\ &= \sqrt{2 + \dfrac{x_1}{x_2} + \dfrac{x_2}{x_1}} \end{align*}By AM-GM, $\dfrac{x_1}{x_2} + \dfrac{x_2}{x_1} > 2$, and the equality case doesn't occur as $x_1 \neq x_2$. Thus, \begin{align*} a_2 &> \sqrt{2+2} \\ &= 2 \end{align*}but $a_2$ is an integer, so we must have $a_2 \geq 3$. $\blacksquare$ Claim 2. $a_n \geq \dfrac{3n-1}{2}$ for all positive odd $n$ up to $2023$, which directly implies the result. Proof. We use induction. The base case is trivial since $a_1 = 1$ for any $x_1$. Now, suppose that $$a_k = \sqrt{(x_1+x_2+\cdots+x_k) \left(\dfrac{1}{x_1} + \dfrac{1}{x_2} + \cdots + \dfrac{1}{x_k}\right)} \geq \dfrac{3k-1}{2}$$or $a_k^2 \geq \dfrac{(3k-1)^2}{4}$ for positive integer $k$. We have \begin{align*} a_{k+2} &= \sqrt{(x_1+x_2+\cdots+x_k + x_{k+1} + x_{k+2}) \left(\dfrac{1}{x_1} + \dfrac{1}{x_2} + \cdots + \dfrac{1}{x_k} + \dfrac{1}{x_{k+1}} + \dfrac{1}{x_{k+2}} \right)} \\ &= \sqrt{(x_1+\cdots+x_k) \left(\dfrac{1}{x_1} + \cdots + \dfrac{1}{x_k} \right) + (x_1 + \cdots + x_k) \left(\dfrac{1}{x_{k+1}} + \dfrac{1}{x_{k+2}} \right) + (x_{k+1}+x_{k+2})\left(\dfrac{1}{x_{k+1}} + \dfrac{1}{x_{k+2}}\right)} \\ &\geq \sqrt{\dfrac{(3k-1)^2}{4} + (x_1 + \cdots + x_k) \left(\dfrac{1}{x_{k+1}} + \dfrac{1}{x_{k+2}}\right) + \dfrac{(x_{k+1}+x_{k+2})(3k-1)^2}{4(x_1+\cdots+x_k)} + 9} \, \, \text{(by the inductive hypothesis and Claim 1)} \\ &\geq \sqrt{\dfrac{(3k-1)^2}{4} + 3(3k-1)+9} \, \, \text{(by Claim 1 and AM-GM)} \\ &= \sqrt{\left(\dfrac{(3k-1)}{2} + 3 \right)^2} \\ &= \dfrac{3k-1}{2} + 3 \\ &= \dfrac{3(k+2)-1}{2} \end{align*}which completes the induction proof. $\blacksquare$ We may finally note that equality holds when $(x_{k+1}+x_{k+2}) \left(\dfrac{1}{x_{k+1}} + \dfrac{1}{x_{k+2}}\right) = 9$, or $x_{k+2} = \dfrac{(7-\sqrt{45})}{2} \cdot x_{k+1}$, and $$(x_1 + \cdots + x_k) \left(\dfrac{1}{x_{k+1}} + \dfrac{1}{x_{k+2}} \right) = \dfrac{(x_{k+1}+x_{k+2})(3k-1)^2}{4(x_1+\cdots+x_k)} + 9$$which is ultimately equivalent to any $x_1, \cdots, x_k$ satisfying $$x_1 + \cdots + x_k = \dfrac{(3k-1)\sqrt{(x_{k+1})(x_{k+2})}}{2}.$$

For brevity, let $$A_k=\sum_{i=1}^{k}x_i, ~~ \text{and} B_k=\sum_{i=1}^{k}\frac{1}{x_i}.$$We will show using induction that $a_n \ge \frac{3n-1}{2}.$ Proof: Base Case: $a_1=\sqrt{x_1\frac{1}{x_1}}=1\ge \frac{3(1)-1}{2}.$ Inductive hypothesis: Assume $a_m \ge \frac{3m-1}{2}$ for $1 \le m\le k.$ Then $$a_{k+1}=\sqrt{\left(A_k+x_{k+1}\right)\left(B_k+\frac{1}{x_{k+1}}\right)}=\sqrt{A_kB_k+1+\frac{A_k}{x_{k+1}}+B_kx_{k+1}}\ge \sqrt{\left(\frac{3k-1}{2}\right)^2+1+\frac{A_k}{x_{k+1}}+B_kx_{k+1}}.$$Let's examine the terms with $x_k+1$ in them. $$\frac{A_k}{x_{k+1}}+B_kx_{k+1}=\sum_{i=1}^k \left(\frac{x_{k+1}}{x_i}+\frac{x_i}{x_{k+1}}\right).$$Claim: For all $1\le i \le n,$ $ \left(\frac{x_{k+1}}{x_i}+\frac{x_i}{x_{k+1}}\right) \ge 7.$ Proof: By the 2-variable case of our inductive hypothesis(strong induction), $a_2 \ge \frac{5}{2}.$ Since $a_2$ is an integer by the problem statement, $a_2 \ge 3.$ Therefore, for pairwise different $x_a,x_b,$ $$\sqrt{(x_a+x_b) \left(\frac{1}{x_a}+\frac{1}{x_b}\right)}\ge 3.$$Squaring both sides and subtracting 2, we show that our claim is true. Therefore, going back to our induction, $$\frac{A_k}{x_{k+1}}+B_kx_{k+1}=\sum_{i=1}^k \left(\frac{x_{k+1}}{x_i}+\frac{x_i}{x_{k+1}}\right)\ge 7k.$$Plugging this back into the expression for $a_{k+1},$ we get $$a_{k+1} \ge \sqrt{\left(\frac{3k-1}{2}\right)^2+1+\frac{A_k}{x_{k+1}}+B_kx_{k+1}}\ge \sqrt{\left(\frac{3k-1}{2}\right)^2+1+7n}=\sqrt{\frac{9k^2-6k+1}{4}+7k+1}=\sqrt{\frac{9k^2-6k+1+28k+4}{4}}$$$$=\sqrt{\frac{9k^2+22k+5}{4}}=\sqrt{\left(\frac{9k^2+12k+4}{4}\right)+\frac{10k+1}{4}} =\sqrt{\left(\frac{3n+2}{2}\right)^2+\frac{10k+1}{4}}>\left(\frac{3n+2}{2}\right).$$Our induction is complete and the claim holds. Now plugging in $n=2023$ shows $a_{2023}\ge \frac{3(2023)-1}{2}=3034.$ Q.E.D.

We show that we can not have $a_n=k$, $a_{n+1}=k+1$, and $a_{n+2}=k+2$, which is sufficient. Let $s=x_1+x_2+\dots+x_n$ and $t=\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}$. Then $$k^2=st$$$$(k+1)^2=(s+x_{n+1})(t+\frac{1}{x_{n+1}})\iff 2k=t\cdot x_{n+1}+s\cdot \frac{1}{x_{n+1}}$$$$(k+2)^2=(s+x_{n+1}+x_{n+2})(t+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})\iff 2k+2=(t+\frac{1}{x_{n+1}})\cdot x_{n+2}+(s+x_{n+1})\cdot \frac{1}{x_{n+1}}$$Applying AM-GM to the third equality we must have that $t\cdot x_{n+1}=s\cdot \frac{1}{x_{n+1}}\iff x_{n+1}=\frac{\sqrt{s}}{\sqrt{t}}$. Applying AM-GM to the fifth equality we must have that $(t+\frac{1}{x_{n+1}})\cdot x_{n+2}=$ $(s+x_{n+1})\cdot \frac{1}{x_{n+1}}\iff x_{n+2}=\frac{\sqrt{s}}{\sqrt{t}}$, a contradiction.

We will show that $a_{n+2} \ge a_n + 3$ which implies the result since $a_1 = 1$. By Cauchy-Schwarz we have \[a_{n+2} = \sqrt{[(x_1+x_2+\dots+ x_n) + (x_{n+1} + x_{n+2})]\left[\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+ \frac 1{x_n} \right)+ \left(\frac{1}{x_{n+1}} + \frac{1}{x_{n+2}}\right)\right]} \ge \sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)} + \sqrt{\frac{(x_{n+1} + x_{n+2})^2 }{x_{n+1}x_{n+2}}} > a_n + 2\]where the final inequality is strict since $x_{n+1} \neq x_{n+2}$ so we have the result $a_{n+2} \ge a_n + 3$ since $a_i$ is an integer for all $i$ so we are done.

Note that $a_1 = 1$ so it suffices to show $a_{n+2}\ge a_n + 3$ for all $n$ and we'll be done by induction. Fix $n$ and let $S = \sum_{i\le n} x_i$ and $T = \sum_{i\le n}1/x_i$. Relabel $x_{n+1}, x_{n+2}$ as $p,q$ for convenience of typesetting. Then note that \[ a_{n+2} = \sqrt{\left(S + p + q\right)\left(T + \frac{1}{p} + \frac{1}{q}\right)}\stackrel{\text{CS}}{\ge} \sqrt{ST} + \sqrt{p\cdot \frac{1}{p}} + \sqrt{q \cdot \frac{1}{q}} = a_n + 2.\]But note that equality only holds if $S/T = p/(1/p) = q/(1/q)$. But this means $p^2 = q^2\implies p = q$, impossible since the $x_i$ are distinct. Hence equality doesn't hold and we have $a_{n+2} > a_n + 2$. Since the $a_i$ are integers, this means $a_{n+2}\ge a_n + 3$ as needed.

Proceed by induction, clearly we have that $a_1=1$, I will now prove $a_{n+2}>a_{n}+2$ which gives $a_{n+2}\geq a_n+3$. We get: \[a_{n+2}^2=a_n^2+\left( \frac{1}{x_{n+1}} +\frac{1}{x_{n+2}} \right)(x_{n+1}+x_{n+2})+(x_{n+1}+x_{n+2})\left(\sum_{i=1}^{n}\frac{1}{x_i}\right)+\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right)\left(\sum_{i=1}^{n}x_i\right)\]Now by two applications of am-gm and the fact $a_{n+1}\neq a_{n+2}$ we get: \[a_{n+2}^2> a_n^2+4+2a_n\sqrt{\left( \frac{1}{x_{n+1}} +\frac{1}{x_{n+2}} \right)(x_{n+1}+x_{n+2})}\]A final am-gm application gives: \[a_{n+2}^2>a_n^2+4a_n+4\]Which suffices.

Intimidating-looking problem that ends up being not so bad at all. We use induction to show $a_{2n-1}\ge 3n-2$ and $a_{2n}\ge 3n$. Base cases $a_1=1$ is trivial, and $a_2\ge 3$ follows from Cauchy-Schwarz $$a_2= \sqrt{(x_1+x_2)\left(\frac{1}{x_1}+\frac{1}{x_2}\right)} > 2\implies a_2\ge 3$$noting that equality case $x_1=x_2$ cannot be obtained. Now assuming $a_{2n-1}\ge 3n-2$ and $a_{2n}\ge 3n$, we first find by Cauchy Schwarz \begin{align*} a_{2n+1} &= \sqrt{(x_1+\cdots +x_{2n}+x_{2n+1})\left(\frac{1}{x_1}+\cdots + \frac{1}{x_{2n}}+\frac{1}{x_{2n+1}}\right)}\\ &= \sqrt{((x_1+\cdots +x_{2n})+x_{2n+1})\left(\frac{a_{2n}^2}{x_1+\cdots +x_{2n}}+\frac{1}{x_{2n+1}}\right)}\\ &\ge a_{2n}+1 \ge 3(n+1)-2 \end{align*}with equality case $x_{2n+1} = \frac{x_1+\cdots +x_{2n}}{a_{2n}}$, as desired. Similarly, we find $a_{2n+2}\ge a_{2n+1}+1$ with equality case $x_{2n+2} = \frac{x_1+\cdots +x_{2n+1}}{a_{2n+1}}$. Now, the key insight is to realize that these two equality cases cannot both be true. If they were, then $x_{2n+1}$ is the average of $x_1, \ldots, x_{2n}$ and $a_{2n}-2n$ zeros. Meanwhile, since for equality case $a_{2n+1} = a_{2n}+1$, then $x_{2n+2}$ is the average of $x_1, \ldots, x_{2n}$ and $a_{2n}-2n$ zeros and $x_{2n+1}$, the average of these numbers, which is still just the average of these numbers. In other words, $x_{2n+2} = x_{2n+1}$, which is disallowed. Thus, in actuality $a_{2n+2}\ge a_{2n+1}+2 \ge 3(n+1)$ as desired. Bringing it back to the problem at hand, we get $a_{2023}\ge 3034$.

This is a really interesting problem, and I was pleasantly surprised when I managed to solve this in contest two years ago. We proceed via induction to show that for all odd integers $n$ , $a_n \ge n + \frac{n-1}{2}$. First note that \[a_1 = \sqrt{x_1 \cdot \frac{1}{x_1}}=1\] Now, assume the claim is true for some odd integer $m=2k+1 \ge1$. Throughout this proof, we will use the following key inequality, Claim : For all positive real numbers $a$ and $b$, \[(a+b)\left(\frac{1}{a}+\frac{1}{b}\right) \ge 4\] Proof : This is simply Cauchy Schwarz. Note that, \[(\sqrt{a}^2 + \sqrt{b}^2)\left(\frac{1}{\sqrt{a}^2}+\frac{1}{\sqrt{b}^2}\right) \ge (1+1)^2 = 4\]as desired. Now, we are ready for the induction. Note, \begin{align*} a_{m+2} &=\sqrt{(x_1+x_2+\dots+x_{m+2})\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_{m+2}}\right)}\\ a_{m+2}^2 & = (x_1+x_2+\dots+x_{m+2})\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_{m+2}}\right)\\ a_{m+2}^2 &= (x_1+x_2+\dots+x_m)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_m}\right) + \left(\frac{1}{x_{m+1}} +\frac{1}{x_{m+2}}\right)(x_{m+1}+x_{m+2})\\ & + (x_1+x_2 + \dots + x_m)\left(\frac{1}{x_{m+1}}+\frac{1}{x_{m+2}}\right) + \left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_m}\right)(x_{m+1}+x_{m+2})\\ & \ge (3k+1)^2 + 4 + \frac{(3k+1)^2(x_{m+1}+x_{m+2})}{x_1+x_2+\dots + x_m} + \frac{4(x_1+x_2+\dots + x_m)}{x_{m+1}+x_{m+2}}\\ & \ge (3k+1)^2 +4 + 2 \sqrt{\frac{4(3k+1)^2(x_{m+1}+x_{m+2})(x_1+x_2+\dots + x_m)}{(x_1+x_2+\dots + x_m)(x_{m+1}+x_{m+2})}}\\ & \ge (3k+1)^2 + 4(3k+1) + 4\\ & \ge (3k+3)^2\\ a_{m+2} & \ge 3k+3 \end{align*}But, if $a_{m+2}=3k+3$, equality must hold at each of the above inequalities. In particular we need \[ \left(\frac{1}{x_{m+1}} +\frac{1}{x_{m+2}}\right)(x_{m+1}+x_{m+2})=4\]which requires $x_{m+1}=x_{m+2}$ which is impossible if $x_1,x_2,\dots,x_{m+2}$ are pairwise distinct positive reals. Thus equality does not hold and $a_{m+2}>3k+3$. Further since $a_{m+2}$ is an integer we must have $a_{m+2} \ge 3k+4$ so, \[a_{m+2} \ge (m+2) + \frac{m+1}{2}\]which completes the induction. Thus, \[a_{2023} \ge 2023 + \frac{2022}{2} = 3034\]

Observe that $a_1 = 1 < a_2 < \dots$. The crux of the problem is the following claim: Claim: There does not exist an index $n$ such that $a_{n+1} - a_n = 1$ and $a_n - a_{n-1} = 1$ simultaneously. Proof: Suppose such an index $n$ existed. Then observe that \begin{align*} a_n^2 &= a_{n-1}^2 + 1 + x_n\left(\frac 1{x_1} + \frac 1{x_2} + \cdots +\frac 1{x_{n-1}}\right) + \frac 1{x_n}\left(x_1+x_2+\cdots+x_n\right) \\ &\geq a_{n-1}^2 + 1 + 2\sqrt{(x_1+x_2+\cdots+x_n)\left(\frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n}\right)} \\ &= (a_n+1)^2. \end{align*}Importantly, equality only holds when $x_n = \sqrt{\frac{x_1+x_2+\cdots+x_{n-1}}{\frac 1{x_1} +\frac 1{x_2}+\cdots+\frac 1{x_{n-1}}}}$. Let $A$ and $B$ be the numerator and denominator of this quantity. The same argument applied to the index $n+1$ yields \[x_{n+1} = \sqrt{\frac{A+\sqrt{\frac AB}}{B+\sqrt{\frac BA}}} = \sqrt{\frac{A\sqrt{AB}+A}{B\sqrt{AB}+B}} = \sqrt{\frac AB} = x_n\]which is a contradiction. $\blacksquare$ So in particular, $a_{n+2} - a_n \geq 3$ for each $n$. It follows that $a_{2023} \geq 1 + 1011 \cdot 3 = 3034$.

Note that $a_1=1$. It suffices to show that $a_{n+2} \geq a_n+3$ for all $n$. Noting that $x_{n+1}, x_{n+2}$ are distinct, \begin{align*} a_{n+2}^2 &=\left(\sum_{i=1}^{n+2} x_i \right)\left(\sum_{i=1}^{n+2} \frac{1}{x_i} \right) \\ &=a_n^2 + (x_{n+1}+x_{n+2})\left(\sum_{i=1}^n \frac{1}{x_i} \right) + \left(\frac{1}{x_{n+1}} + \frac{1}{x_{n+2}} \right)\left(\sum_{i=1}^n x_i \right) + (x_{n+1}+x_{n+2})\left(\frac{1}{x_{n+1}} + \frac{1}{x_{n+2}} \right) \\ &> a_n^2 + (x_{n+1}+x_{n+2})\left(\sum_{i=1}^n \frac{1}{x_i} \right) + \left(\frac{4}{x_{n+1}+x_{n+2}} \right)\left(\sum_{i=1}^n x_i \right) + 4 \\ &\geq a_n^2 + 4 \sqrt{\left(\sum_{i=1}^n \frac{1}{x_i} \right)\left(\sum_{i=1}^n x_i \right)} + 4 =(a_n+2)^2. \end{align*}Since $a_{n+2}$ and $a_n$ are integers, $a_{n+2} > a_n+2 \implies a_{n+2} \geq a_n+3$ so we are done.

Very beautiful problem, definitely one of the nicer algebra problems out there. Claim 1: $a_{n+1} \ge a_n + 1$ with equality iff $x_{n+1} = \frac{x_1 + \dots + x_n}{a_n}$. Proof: We have $$a_{n+1} = \sqrt{a_n^2 + 1 + x_{n+1} \left(\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}\right) +\frac{ x_1 + x_2 + ... + x_n}{x_{n+1}}}$$$$\stackrel{\textrm{AM-GM}}{\ge} \sqrt{a_n^2 + 1 + 2a_n} = a_n + 1.$$Equality holds iff equality holds in AM-GM, i. e. $$x_{n+1}^2 = \frac{x_1 + x_2 + ... + x_n}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}} = \frac{ x_1 + x_2 + ... + x_n}{\frac{a_n}{ x_1 + x_2 + ... + x_n}}$$$$\iff x_{n+1} = \frac{x_1 + \dots + x_n}{a_n}$$as desired. $\square$ Claim 2: We cannot simultaneously have $a_{n+2} = a_{n+1} + 1, a_{n+1} = a_n + 1$. Proof: Note that for equality to hold in both cases, we need $$x_{n+2} = \frac{x_1 + x_2 + \dots + x_n + x_{n+1}}{a_n + 1} = \frac{\frac{a_n+1}{a_n} \times x_1 + \dots + x_n}{a_n+1} = \frac{x_1 + x_2 + \dots + x_n}{a_n} = x_{n+1}.$$But the sequence is composed of distinct terms, so this is impossible. $\square$ Note that due to Claim 2, we must have $a_{n+2} \ge a_n + 3$. Therefore, $$a_{2023} \ge a_{2021} + 3$$$$\ge a_{2019} + 6$$$$\ge a_{2017} + 9$$$$\vdots$$$$\ge a_1 + 3 \times 1011 = 3034$$as desired. $\blacksquare$

Let $S_n := x_1 + \cdots + x_n$ and $T_n := \frac{1}{x_1} + \cdots + \frac{1}{x_n}$, then \[\begin{aligned} a_{n + 2}^2 &= (S_n + x_{n + 1} + x_{n + 2})\left(T_n + \frac{1}{x_{n + 1}} + \frac{1}{x_{n + 2}}\right) \\ &= S_n T_n + \frac{S_n}{x_{n + 1}} + x_{n + 1}T_n + \frac{S_n}{x_{n + 2}} + x_{n + 2}T_n + \frac{x_{n+2}}{x_{n+1}} + \frac{x_{n + 1}}{x_{n+2}} + 2 \\ &\ge a_n^2 + 2a_n + 2a_n + 4 = (a_n + 2)^2 \end{aligned}\] The equality cannot hold because $x_{n + 1} \neq x_{n + 2}$, thus $a_{n + 2} \ge a_n + 3$ and $a_{2n + 1} \ge 3n + 1$ follows easily by induction. I am surprised that it is rated A3 on the shortlist, considering the difficulty of this problem.