Problem 4: 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 \(a_{2023} \geqslant 3034.\)
Hint: 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 \(a_{2023}\geq a_{2021}+3\geq\cdots\geq a_1+3\times 1011=3034.\blacksquare\)
Solution 1: 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 \(a_{2023} \ge a_1+3\cdot 1011 = 3034\) . \(\square\)
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 \(a_{n-1}<a_{n}\)
\[2a_{n-1}=\frac{s}{x_n}+xh\] \[h=\frac{a^2}{s}\]2 \(a_{n-1}=\frac{s}x+\frac{x}s{a_{n-1}^2}\)
let \(k = \frac{s}x\)
\(2a_{n-1}\) =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 \(a_{n-1}+3.\)