Kvant Math Problem 1432

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Problem

Prove that for any sequence of positive numbers $a_n$, the integer parts of the square roots of the numbers $$b_n=(a_1+a_2+\ldots+a_n)\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\ldots+\dfrac{1}{a_n}\right)$$ are all distinct.

L. D. Kurylandchik

Exploration

Consider a sequence of positive numbers $a_1, a_2, \dots$. Define

$b_n = (a_1 + \dots + a_n)\left(\frac{1}{a_1} + \dots + \frac{1}{a_n}\right).$

Compute small cases to understand its behavior. For $n=1$, $b_1 = a_1 \cdot \frac{1}{a_1} = 1$. For $n=2$, if $a_1 = 1$, $a_2 = 1$, then $b_2 = (1+1)(1 + 1) = 4$. If $a_1 = 1$, $a_2 = 2$, then $b_2 = (1+2)(1 + 1/2) = 3 \cdot 1.5 = 4.5$, integer part $4$. For $n=3$, choose $a_1 = a_2 = a_3 = 1$, then $b_3 = 3 \cdot 3 = 9$. Choosing $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, we have $b_3 = 6 \cdot (1 + 1/2 + 1/3) = 6 \cdot \frac{11}{6} = 11$, integer part $11$.

The integer parts grow and appear distinct. The quantity $b_n$ resembles a discrete Cauchy–Schwarz inequality: $(a_1 + \dots + a_n)(1/a_1 + \dots + 1/a_n) \ge n^2$, equality only if all $a_i$ are equal. This suggests $b_n$ grows and cannot return to a previous integer.

The main difficulty is to prove that no two numbers $b_m$ and $b_n$ share the same integer part for $m \neq n$, regardless of the choice of positive $a_i$.

Problem Understanding

The problem asks to show that for any sequence of positive numbers $a_n$, the integer parts $\lfloor \sqrt{b_n} \rfloor$ of the sequence

$b_n = (a_1 + \dots + a_n)\left(\frac{1}{a_1} + \dots + \frac{1}{a_n}\right)$

are all distinct. This is a Type B problem: we are asked to prove a statement without classifying or optimizing.

The core difficulty lies in showing that the sequence grows fast enough relative to its predecessor so that the integer part of the square root increases, and no integer is repeated. The crucial insight is the inequality $b_{n+1} > b_n$, but we must also prove a stronger inequality to guarantee that the integer parts never coincide. The Cauchy–Schwarz inequality provides a lower bound $b_n \ge n^2$, which may help in separating the integer parts.

Proof Architecture

Lemma 1: For any sequence of positive numbers $a_1, \dots, a_n$,

$b_n \ge n^2,$

with equality if and only if all $a_i$ are equal. This follows from the Cauchy–Schwarz inequality applied to $(\sqrt{a_1}, \dots, \sqrt{a_n})$ and $(1/\sqrt{a_1}, \dots, 1/\sqrt{a_n})$.

Lemma 2: For any positive number $a_{n+1}$,

$b_{n+1} > b_n.$

This follows from considering the increment $(a_1 + \dots + a_n + a_{n+1})(1/a_1 + \dots + 1/a_n + 1/a_{n+1}) - (a_1 + \dots + a_n)(1/a_1 + \dots + 1/a_n)$ and using positivity of $a_{n+1}$ and the convexity-like structure.

Lemma 3: If $b_{n+1} > b_n$, then $\lfloor \sqrt{b_{n+1}} \rfloor \ge \lfloor \sqrt{b_n} \rfloor$. This is trivial from the monotonicity of the floor function.

Lemma 4 (Key Lemma): The increment $b_{n+1} - b_n < 2\sqrt{b_n} + 1$ is impossible. Equivalently, the integer part of the square root cannot stay constant, because the distance between consecutive squares is at least $2\lfloor \sqrt{b_n} \rfloor + 1$, which ensures distinct integer parts. This requires careful analysis of the growth of $b_n$.

The hardest step is Lemma 4, as it involves showing that the growth is sufficient to prevent a repeated integer part for arbitrary positive sequences.

Solution

We first establish Lemma 1. By Cauchy–Schwarz,

$(a_1 + \dots + a_n)\left(\frac{1}{a_1} + \dots + \frac{1}{a_n}\right) \ge \left(\sqrt{a_1 \cdot \frac{1}{a_1}} + \dots + \sqrt{a_n \cdot \frac{1}{a_n}}\right)^2 = n^2,$

with equality only if $\sqrt{a_i} / (1/\sqrt{a_i})$ is constant for all $i$, i.e., $a_1 = \dots = a_n$. Therefore $b_n \ge n^2$.

To prove Lemma 2, write

$$b_{n+1} - b_n = (a_1 + \dots + a_n + a_{n+1})\left(\sum_{i=1}^{n} \frac{1}{a_i} + \frac{1}{a_{n+1}}\right) - (a_1 + \dots + a_n)\sum_{i=1}^{n} \frac{1}{a_i}.$$

Expanding, we get

$$b_{n+1} - b_n = a_{n+1}\sum_{i=1}^{n} \frac{1}{a_i} + (a_1 + \dots + a_n)\frac{1}{a_{n+1}} + a_{n+1}\frac{1}{a_{n+1}} = a_{n+1}\sum_{i=1}^{n} \frac{1}{a_i} + \frac{a_1 + \dots + a_n}{a_{n+1}} + 1.$$

Since $a_i > 0$, each term is positive, hence $b_{n+1} - b_n > 0$, proving Lemma 2.

Lemma 3 follows from the monotonicity of the floor function and the fact that $b_{n+1} > b_n$ implies $\sqrt{b_{n+1}} > \sqrt{b_n}$, hence $\lfloor \sqrt{b_{n+1}} \rfloor \ge \lfloor \sqrt{b_n} \rfloor$.

To prove Lemma 4, suppose by contradiction that $\lfloor \sqrt{b_{n+1}} \rfloor = \lfloor \sqrt{b_n} \rfloor = k$ for some integer $k$. Then $k^2 \le b_n < (k+1)^2$ and $k^2 \le b_{n+1} < (k+1)^2$, which implies $b_{n+1} - b_n < (k+1)^2 - k^2 = 2k + 1$.

However, from the previous expansion,

$$b_{n+1} - b_n = a_{n+1}\sum_{i=1}^{n} \frac{1}{a_i} + \frac{a_1 + \dots + a_n}{a_{n+1}} + 1 \ge 2 \sqrt{(a_1 + \dots + a_n)\sum_{i=1}^{n} \frac{1}{a_i}} + 1 = 2 \sqrt{b_n} + 1,$$

by the arithmetic–geometric mean inequality applied to the two positive numbers $a_{n+1}\sum_{i=1}^{n} 1/a_i$ and $(a_1 + \dots + a_n)/a_{n+1}$. Since $b_n \ge n^2 \ge 1$, it follows that $2\sqrt{b_n} + 1 > 2k + 1$, a contradiction. Therefore the integer part must increase.

As a result, the integer parts $\lfloor \sqrt{b_n} \rfloor$ strictly increase with $n$ and are therefore distinct.

This completes the proof.

∎

Verification of Key Steps

The crucial step is the application of the arithmetic–geometric mean inequality to the increment:

$$a_{n+1}\sum_{i=1}^{n} \frac{1}{a_i} + \frac{a_1 + \dots + a_n}{a_{n+1}} \ge 2\sqrt{(a_1 + \dots + a_n)\sum_{i=1}^{n} \frac{1}{a_i}}.$$

Check a concrete case: $n=2$,