Problem 4

Let

x_1,x_2,\ldots,x_{2023}

be pairwise different positive real numbers such that

a_n=(x_1+x_2+\cdots+x_n)\left(\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n}\right)

is an integer for every

n=1,2,\ldots,2023

. Prove that

a_{2023}\ge3034

Step 3 of 3: Strict AM-GM forces a jump of at least 3

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In plain words

Since the two chosen numbers are always distinct, the ratio u can never equal 1, so the bound from Step 2 is strictly more than a_n+2, and integrality then upgrades this to a genuine increase of at least 3 every two steps.

a_{2m+1}\ge3m+1\implies a_{2023}\ge3034

Detailed analysis

Because $x_{n+1}\ne x_{n+2}$ , $u\ne1$ , so by AM-GM $u+\frac1u>2$ strictly, giving $a_{n+2}>a_n+2$ from Step 2. Since $a_n,a_{n+2}$ are both integers, $a_{n+2}-a_n$ is an integer exceeding $2$ , hence $a_{n+2}\ge a_n+3$ . Combined with $a_1=1$ (Step 1), induction on $m$ gives $a_{2m+1}\ge3m+1$ for every $m\ge0$ . Taking $m=1011$ (so $2m+1=2023$ ) yields $a_{2023}\ge3\cdot1011+1=3034$ .