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 2 of 3: A three-term Cauchy-Schwarz step

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In plain words

Splitting the sum defining $a_{n+2}$ into the first n terms plus the two new terms and applying Cauchy-Schwarz with three matched pairs produces a lower bound built from $a_n$ and the ratio of the two new variables.

a_{n+2}\ge\left(\sqrt{a_n}+u+\dfrac1u\right)^2\ge a_n+u+\dfrac1u,\qquad u=\sqrt{\dfrac{x_{n+1}}{x_{n+2}}}

Detailed analysis

Write $A=x_1+\cdots+x_n$ and $B=\frac1{x_1}+\cdots+\frac1{x_n}$ , so $a_n=AB$ . The three-term Cauchy-Schwarz inequality gives $\sqrt{a_{n+2}}=\sqrt{(A+x_{n+1}+x_{n+2})\left(B+\frac1{x_{n+1}}+\frac1{x_{n+2}}\right)}\ge\sqrt{AB}+\sqrt{\frac{x_{n+1}}{x_{n+2}}}+\sqrt{\frac{x_{n+2}}{x_{n+1}}}=\sqrt{a_n}+u+\frac1u$ , where $u=\sqrt{x_{n+1}/x_{n+2}}$ . Squaring gives $a_{n+2}\ge(\sqrt{a_n}+u+1/u)^2\ge a_n+u+1/u$ .