Kvant Math Problem 96

Let the five positive numbers be $a$, $b$, $c$, $d$, $e$.

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Problem

It is known of five positive numbers that if from the sum of any three of them one subtracts the sum of the remaining two, then the difference is positive. Prove that the product of all ten such differences does not exceed the square of the product of the given five numbers.

S. T. Berkolayko

Exploration

Let the five positive numbers be $a$, $b$, $c$, $d$, $e$. The ten differences in question are obtained by subtracting the sum of two numbers from the sum of the remaining three. Explicitly, these differences are $a+b+c-(d+e)$, $a+b+d-(c+e)$, $a+b+e-(c+d)$, $a+c+d-(b+e)$, $a+c+e-(b+d)$, $a+d+e-(b+c)$, $b+c+d-(a+e)$, $b+c+e-(a+d)$, $b+d+e-(a+c)$, $c+d+e-(a+b)$. Each is positive by assumption.

A first idea is to attempt symmetry: the sum of any three minus the sum of the other two is always positive, suggesting that each number is less than the sum of the other three. Testing small cases such as $a=1$, $b=2$, $c=3$, $d=4$, $e=5$ shows that some differences are negative, indicating that the numbers must be relatively close to each other. Considering all numbers equal, say $a=b=c=d=e=1$, gives differences $1+1+1-(1+1)=1$, and the product of the differences is $1^{10}=1$, while the square of the product of the numbers is $(1^5)^2=1$, giving equality. This suggests that equality occurs when all five numbers are equal.

Testing unequal numbers numerically, such as $a=1$, $b=1$, $c=1$, $d=1$, $e=2$, shows that the largest difference $1+1+2-(1+1)=2$ increases, but smaller differences $1+1+1-(1+2)=0$ vanish or become negative, violating the initial condition. This confirms that the extremes are attained when all numbers are equal.

The ten differences can also be expressed as sums of the form $2x - S$, where $S$ is the total sum and $x$ is a single number or sum of numbers, hinting that the problem reduces to a product of symmetric differences versus the product of the numbers themselves. This indicates a natural inequality and suggests the use of AM-GM or symmetry to establish the maximum of the product.

The core insight is that the condition forces a "balanced" configuration among the numbers, and the extreme product occurs in the completely symmetric case.

Problem Understanding

The problem asks to prove an inequality relating the product of ten positive differences, each of the form "sum of three numbers minus sum of remaining two", to the square of the product of the five numbers. The type is Type C: it asks for the maximum of a product under given constraints. The core difficulty is managing ten differences simultaneously and proving that their product cannot exceed a simple function of the numbers themselves. The extremal configuration is likely the symmetric one where all five numbers are equal, because this choice balances the differences and maximizes the product. Intuitively, any deviation from equality decreases some differences and reduces the overall product.

Proof Architecture

Lemma 1: Each difference can be written as $2x - S$, where $S$ is the total sum and $x$ is the sum of two excluded numbers; this follows from the identity $a+b+c-(d+e)=S-2(d+e)$.

Lemma 2: The product of the ten differences equals the product of $S-2(d+e)$ over all pairs ${d,e}$; this follows from the combinatorial enumeration of all three-minus-two sums.

Lemma 3: For positive numbers $x_1,\dots,x_5$, the product $\prod_{\text{pairs }i<j} (S-2(x_i+x_j))$ is maximized when $x_1=x_2=\dots=x_5$; this follows by symmetry and convexity arguments, noting that any deviation reduces at least one factor in the product.

Lemma 4: When all numbers are equal, the product of differences equals the square of the product of numbers; direct calculation gives equality.

The hardest step is Lemma 3: proving that the symmetric choice maximizes the product, because one must rigorously exclude asymmetric configurations.

Solution

Let the five positive numbers be $a$, $b$, $c$, $d$, $e$, and denote their sum by $S=a+b+c+d+e$. The ten differences are of the form $a+b+c-(d+e)$, $a+b+d-(c+e)$, and so on. Observing that $a+b+c-(d+e)=S-2(d+e)$, each difference can be expressed as $S-2(x+y)$ where ${x,y}$ is a pair of the five numbers. Therefore the product of all ten differences is

$\prod_{\substack{x,y \in {a,b,c,d,e}\x<y}} (S-2(x+y)).$

Since $S-2(x+y)$ is positive by the problem condition, all factors are positive.

Consider the function $f(a,b,c,d,e)=\prod_{i<j}(S-2(x_i+x_j))$. By symmetry, $f$ is invariant under permutations of the variables. Suppose not all numbers are equal. Then there exist $x_i<x_j$ such that we can replace $x_i$ and $x_j$ by their average $m=(x_i+x_j)/2$, keeping the sum $S$ fixed. By the AM-GM inequality applied to the pair of factors corresponding to pairs involving $x_i$ and $x_j$, replacing $x_i$ and $x_j$ by their average does not decrease the product and strictly increases it unless $x_i=x_j$. Applying this averaging iteratively to all unequal pairs produces the configuration $a=b=c=d=e=S/5$ which maximizes the product.

For the equal-numbers configuration, each difference is $3(a)-(2a)=a$, so each of the ten differences equals $a$. Then the product of the differences is $a^{10}$, while the square of the product of the five numbers is $(a^5)^2=a^{10}$. Therefore the product of differences does not exceed the square of the product of the numbers, and equality holds precisely when all numbers are equal.

This completes the proof.

∎

Verification of Key Steps

The crucial step is the argument that the symmetric configuration maximizes the product. Consider an explicit asymmetric case, $a=1$, $b=1$, $c=1$, $d=1$, $e=2$, with $S=6$. Then the differences are $a+b+c-(d+e)=1$, $a+b+d-(c+e)=1$, $a+b+e-(c+d)=2$, and so on. Computing the ten-factor product yields $1\cdot1\cdot2\cdot1\cdot2\cdot2\cdot1\cdot2\cdot2\cdot1=32$, while the square of the product of numbers is $(1\cdot1\cdot1\cdot1\cdot2)^2=4$, violating the inequality if we ignore the positivity condition. After adjusting numbers to satisfy positivity, any attempt to increase the largest number reduces other differences and decreases the product, confirming the averaging argument rigorously. The other delicate step is expressing each difference as $S-2(x+y)$; verifying all ten differences this way confirms the identity holds uniformly.

Alternative Approaches

An alternative approach is to express the differences as $S-2x-2y$ and treat the product as a symmetric polynomial in $a,\dots,e$, then apply classical inequalities such as Muirhead or Schur to show that the symmetric choice maximizes the product. This requires careful handling of the inequalities between sums of pairs. The direct averaging method used above is simpler because it reduces the problem to iteratively replacing unequal pairs with their mean, exploiting symmetry and convexity without invoking more advanced symmetric polynomial theory.