IMO 2023 Problem 4

22 Sep, 2024

Problem

Let x1, x2, . . . , x2023 be pairwise different positive real numbers such that

a_{n} = (x_{1} + x_{2} + \dots + x_{n}) (\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{n}})

is an integer for every $n = 1, 2, . . ., 2023$ . Prove that $a_{2023} ⩾ 3034$ .

Conclusion

Here we prove for a strengthened conclusion, for $k \in N +$

\forall n = 2 k + 1, a_{n} \geq 3 k + 1

Proof by induction

For $n = 2$ , accourding to Cauchy's inequality, we have

a_{2} = \sqrt{(x_{1} + x_{2}) (\frac{1}{x_{1}} + \frac{1}{x_{2}})} \geq 2

equality holds only when $x_{1} = x_{2}$ . Given the condition that $x_{i}$ are distinct, we have $a_{2} > 2$ . Since $a_{2}$ is an integer, we have $a_{2} \geq 3$ .

For $k = 1, n = 3$ ,

a_{3} = \sqrt{(x_{1} + x_{2} + x_{3}) (\frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}})}

= \sqrt{(x_{1} + x_{2}) (\frac{1}{x_{1}} + \frac{1}{x_{2}}) + x_{3} (\frac{1}{x_{1}} + \frac{1}{x_{2}}) + \frac{1}{x_{3}} (x_{1} + x_{2}) + 1}

= \sqrt{a_{2}^{2} + x_{3} \frac{a_{2}^{2}}{x_{1} + x_{2}} + \frac{1}{x_{3}} (x_{1} + x_{2}) + 1}

\geq \sqrt{9 + x_{3} \frac{9}{x_{1} + x_{2}} + \frac{1}{x_{3}} (x_{1} + x_{2}) + 1}

= \sqrt{10 + 9 \frac{x_{3}}{x_{1} + x_{2}} + \frac{x_{1} + x_{2}}{x_{3}}}

\geq 4 = 3 k + 1

equality holds only when $x_{1} + x_{2} = 3 x_{3}$ , which is plausible.

Suppose the claim holds for $k$ that $a_{2 k + 1} \geq 3 k + 1$ , then for $k + 1$ , we have

a_{2 k + 2} = \sqrt{(x_{1} + x_{2} + \dots + x_{2 k + 2}) (\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{2 k + 2}})}

= \sqrt{(x_{1} + x_{2} + \dots + x_{2 k + 1}) (\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{2 k + 1}}) + x_{2 k + 2} (\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{2 k + 1}}) + \frac{1}{x_{2 k + 2}} (x_{1} + x_{2} + \dots + x_{2 k + 1}) + 1}

= \sqrt{a_{2 k + 1}^{2} + x_{2 k + 2} \frac{a_{2 k + 1}^{2}}{x_{1} + x_{2} + \dots + x_{2 k + 1}} + \frac{1}{x_{2 k + 2}} (x_{1} + x_{2} + \dots + x_{2 k + 1}) + 1}

\geq \sqrt{(3 k + 1)^{2} + x_{2 k + 2} \frac{{(3 k + 1)}^{2}}{x_{1} + x_{2} + \dots + x_{2 k + 1}} + \frac{1}{x_{2 k + 2}} (x_{1} + x_{2} + \dots + x_{2 k + 1}) + 1}

\geq \sqrt{(3 k + 1)^{2} + 2 (3 k + 1) + 1} = 3 k + 2

equality holds only when

x_{1} + x_{2} + \dots + x_{2 k + 1} = (3 k + 1) x_{2 k + 2}

For $a_{2 k + 3} = a_{2 (k + 1) + 1}$

a_{2 k + 3} = \sqrt{(x_{1} + x_{2} + \dots + x_{2 k + 2}) (\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{2 k + 2}}) + x_{2 k + 3} (\frac{1}{x_{1}} + \frac{1}{x_{2}} + \dots + \frac{1}{x_{2 k + 2}}) + \frac{1}{x_{2 k + 3}} (x_{1} + x_{2} + \dots + x_{2 k + 2}) + 1}

= \sqrt{a_{2 k + 2}^{2} + x_{2 k + 3} \frac{a_{2 k + 2}^{2}}{x_{1} + x_{2} + \dots + x_{2 k + 2}} + \frac{1}{x_{2 k + 3}} (x_{1} + x_{2} + \dots + x_{2 k + 2}) + 1}

\geq \sqrt{(3 k + 2)^{2} + x_{2 k + 3} \frac{{(3 k + 2)}^{2}}{x_{1} + x_{2} + \dots + x_{2 k + 2}} + \frac{1}{x_{2 k + 3}} (x_{1} + x_{2} + \dots + x_{2 k + 2}) + 1}

\geq \sqrt{(3 k + 2)^{2} + 2 (3 k + 2) + 1} = 3 k + 3

equality holds only when

x_{1} + x_{2} + \dots + x_{2 k + 2} = (3 k + 2) x_{2 k + 3}

The two conditions above cannot be satisfied at the same time, otherwise we have

x_{1} + x_{2} + \dots + x_{2 k + 1} + x_{2 k + 2} = (3 k + 1) x_{2 k + 2} + x_{2 k + 2} = (3 k + 2) x_{2 k + 3} \Rightarrow x_{2 k + 2} = x_{2 k + 3}

which is impossible.

Therefore, since both equalities cannot be satisfied at the same time, we have

a_{2 k + 3} > 3 k + 2 \Rightarrow a_{2 k + 3} \geq 3 k + 3

#math-competition