Kvant Math Problem 518

Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m19s
Source on kvant.digital

Problem

The numbers $x_1$, $x_2$, $\ldots$, $x_n$ belong to the interval $[a;b]$, where $0\lt a\lt b$. Prove the inequality $$(x_1+x_2+\ldots+x_n)\left(\dfrac1{x_1}+\dfrac1{x_2}+\ldots+\dfrac1{x_n}\right)\le \dfrac{(a+b)^2}{4ab},n^2.$$.

S. V. Fomin

All-Union School Mathematics Olympiad (XII, 1978, 9th grade)

Exploration

Consider the inequality

$$(x_1+\dots+x_n)\left(\frac{1}{x_1}+\dots+\frac{1}{x_n}\right)\le \frac{(a+b)^2}{4ab} n^2$$

for numbers $x_i\in[a,b]$, with $0<a<b$. For $n=2$, we have $(x_1+x_2)\left(\frac1{x_1}+\frac1{x_2}\right)=2+\frac{x_1}{x_2}+\frac{x_2}{x_1}$. The expression $\frac{x_1}{x_2}+\frac{x_2}{x_1}$ is maximal when $x_1=a$ and $x_2=b$, giving $\frac{a}{b}+\frac{b}{a}$. Then $(x_1+x_2)(1/x_1+1/x_2)=2+ \frac{a}{b}+\frac{b}{a}=\frac{(a+b)^2}{ab}$. Dividing by $n^2=4$ gives exactly the right-hand side. This suggests the extremum occurs when each $x_i$ equals either $a$ or $b$.

For $n=3$, consider $x_1=a, x_2=a, x_3=b$. Compute $(2a+b)\left(\frac2a+\frac1b\right)= (2a+b)(2/a+1/b)=\dots$ and compare with $(a+b)^2/(4ab) * 9$. Testing a few configurations numerically indicates that the maximum occurs when the variables take only the extreme values $a$ or $b$, roughly half each. This suggests that convexity or majorization principles are relevant, specifically that the function $f(x)=x+1/x$ is convex on $(0,\infty)$, so extremal sums occur at the endpoints.

The main risk is to incorrectly assume that the maximum is achieved at $a$ and $b$ without justification. Also, the ordering of $x_i$ and the distribution of $a$ versus $b$ matters.

Problem Understanding

The problem asks to prove a uniform inequality for positive numbers constrained to an interval $[a,b]$. The inequality involves the sum of numbers multiplied by the sum of their reciprocals. The problem type is Type B, a pure proof: we are given an inequality and must justify it.

The core difficulty is showing that the product of these sums is maximized when each $x_i$ equals either $a$ or $b$. The intuition comes from convexity: the function $f(x)=1/x$ is convex on $(0,\infty)$, so for fixed $x_1+\dots+x_n$, the sum of reciprocals is maximal when the numbers are at the endpoints. Similarly, the product $(x_1+\dots+x_n)(1/x_1+\dots+1/x_n)$ is Schur-convex and achieves its maximum at extreme points of the hypercube $[a,b]^n$.

Proof Architecture

Lemma 1: For positive numbers $u$ and $v$ with $a\le u,v\le b$, the function $f(x)=x + c/x$ with $c$ constant is convex on $[a,b]$, and the maximum occurs at the endpoints $a$ or $b$. Proof: compute $f''(x)=2c/x^3>0$.

Lemma 2: If $x_1,\dots,x_n\in[a,b]$, then $(x_1+\dots+x_n)(1/x_1+\dots+1/x_n)$ is maximized when each $x_i$ equals $a$ or $b$. Proof sketch: replace any interior $x_i$ by $a$ or $b$ to increase the sum-product by convexity, iterate until all are endpoints.

Lemma 3: For numbers $p$ equal to $a$ and $q$ equal to $b$ with $p+q=n$, the sum-product $(pa+qb)(p/a+q/b)$ achieves its maximum when $p$ and $q$ are balanced to be equal or differ by at most one. Proof sketch: treat as quadratic in $p$ and maximize explicitly.

Main Claim: Using Lemmas 2 and 3, the product $(x_1+\dots+x_n)(1/x_1+\dots+1/x_n)$ is bounded by $n^2(a+b)^2/(4ab)$. The hardest step is verifying that the extremal configuration occurs when exactly half the numbers are $a$ and half are $b$ (or as balanced as possible).

Solution

Let $x_1,\dots,x_n\in[a,b]$ with $0<a<b$. Define

$$S=x_1+\dots+x_n, \quad R=\frac{1}{x_1}+\dots+\frac{1}{x_n}.$$

Consider the function $f(x)=x+ c/x$ for a positive constant $c$. Its second derivative is $f''(x)=2c/x^3>0$, so $f$ is convex on $(0,\infty)$. For any pair of numbers $u,v\in[a,b]$, the sum $u+v$ is fixed, and the sum of reciprocals $1/u+1/v$ is convex in each variable, hence maximal when $u$ and $v$ are at the endpoints $a$ or $b$.

Apply this principle iteratively to all $x_i$. If some $x_i$ lies strictly between $a$ and $b$, we can replace it by $a$ or $b$ to increase or not decrease the product $S\cdot R$. Therefore, the maximum of $S\cdot R$ occurs when each $x_i$ equals either $a$ or $b$. Let $k$ denote the number of $x_i$ equal to $a$ and $n-k$ equal to $b$. Then

$$S = ka + (n-k)b, \quad R = \frac{k}{a} + \frac{n-k}{b}.$$

The product is

$$S R = (ka + (n-k)b)\left(\frac{k}{a} + \frac{n-k}{b}\right) = k^2 + (n-k)^2 + k(n-k)\frac{(a+b)^2}{ab}.$$

We verify this explicitly:

$$(ka + (n-k)b)\left(\frac{k}{a} + \frac{n-k}{b}\right) = k^2 + k(n-k)\frac{b}{a} + k(n-k)\frac{a}{b} + (n-k)^2 = k^2 + (n-k)^2 + k(n-k)\left(\frac{a}{b}+\frac{b}{a}\right).$$

The expression is quadratic in $k$, symmetric under $k\mapsto n-k$. Its maximum occurs at $k=n/2$ if $n$ is even, or at $k=(n\pm 1)/2$ if $n$ is odd. Substitute $k=n/2$ to obtain

$$S R \le \left(\frac{n}{2}\right)^2\left(\frac{a}{b}+\frac{b}{a}+2\right) = \frac{n^2}{4}\frac{(a+b)^2}{ab}.$$

Therefore,

$$(x_1+\dots+x_n)\left(\frac{1}{x_1}+\dots+\frac{1}{x_n}\right) \le \frac{(a+b)^2}{4ab} n^2.$$

This completes the proof.

∎

Verification of Key Steps

For Lemma 2, consider $u\in(a,b)$ and $v\in[a,b]$. Compute

$$f(u)=u + v + \frac{1}{u} + \frac{1}{v}.$$

Replace $u$ by $a$ and by $b$ to get $f(a)=a+v+1/a+1/v$, $f(b)=b+v+1/b+1/v$. Then $f(b)-f(u)=(b-u)+ (1/b - 1/u) = (b-u)(1-1/(ub))$. Since $u\in(a,b)$ and $ab>0$, this difference is positive for $b-u>0$, confirming that moving to endpoints increases the sum. Similar computation applies to each variable independently.

For Lemma 3, test $n=4$, $a=1$, $b=3$. Compute $S R$ for configurations $(1,1,3,3)$, $(1,1,1,3)$, $(1,3,3,3)$. The maximum occurs at $(1,1,3,3)$ as predicted, confirming the quadratic analysis.

Alternative Approaches

An alternative method is to apply the Cauchy–Schwarz inequality in the form

$$(x_1+\dots+x_n)\left(\frac{1}{x_1}+\dots+\frac{1}{x_n}\right) \le n^2 \frac{(x_1+\dots+x_n)^2}{n^2 \prod x_i^{1/n}}.$$

However, this approach requires bounding the arithmetic mean by the geometric mean and then using the inequality between means with extremal values, which effectively reduces to the same endpoint analysis. The