IMO 1972 Problem 4
The system involves five positive real numbers $(x_1, x_2, x_3, x_4, x_5)$ linked cyclically by inequalities of the form $(x_i^2 - x_{i+2}x_{i+4})(x_{i+1}^2 - x_{i+2}x_{i+4}) \le 0$, where indices are…
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 6m59s
Problem
Find all solutions $(x_1, x_2, x_3, x_4, x_5)$ of the system of inequalities $$ (x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \leq 0 \ ,(x_2^2 - x_4x_1)(x_3^2 - x_4x_1) \leq 0 \ ,(x_3^2 - x_5x_2)(x_4^2 - x_5x_2) \leq 0 \ ,(x_4^2 - x_1x_3)(x_5^2 - x_1x_3) \leq 0 \ ,(x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \leq 0 $$ where $x_1, x_2, x_3, x_4, x_5$ are positive real numbers.
Exploration
The system involves five positive real numbers $(x_1, x_2, x_3, x_4, x_5)$ linked cyclically by inequalities of the form $(x_i^2 - x_{i+2}x_{i+4})(x_{i+1}^2 - x_{i+2}x_{i+4}) \le 0$, where indices are taken modulo $5$. Each inequality demands that one of two consecutive terms, $x_i^2 - x_{i+2}x_{i+4}$ or $x_{i+1}^2 - x_{i+2}x_{i+4}$, is nonpositive and the other nonnegative, so that their product is nonpositive. The cyclic pattern suggests symmetry may be exploitable. A natural first guess is that all $x_i$ are equal, which indeed satisfies the system as each term $x_i^2 - x_{i+2}x_{i+4}$ becomes zero, giving products zero. Another possibility is a configuration with two distinct values alternating, but a small test shows that for the five-cycle, such alternation is inconsistent. Testing explicit small ratios also indicates that any deviation from equality leads to contradictions in the cyclic chain of inequalities. The key difficulty is ensuring consistency across the full cycle of five inequalities; naive local adjustments immediately violate another inequality. This hints that the solution set may be the completely symmetric one $x_1 = x_2 = x_3 = x_4 = x_5$.
Problem Understanding
The problem is Type A, a classification problem, since it asks for all positive real solutions of a system of inequalities. The objects are positive real numbers arranged cyclically, and the task is to determine all quintuplets $(x_1, x_2, x_3, x_4, x_5)$ that satisfy the five given inequalities simultaneously. The difficulty lies in the interplay of cyclic dependencies; an inequality involving $x_i, x_{i+1}, x_{i+2}, x_{i+4}$ constrains local differences, but any local variation propagates around the cycle. A plausible solution set is the uniform tuple where all five numbers are equal, because in that case all inequalities reduce to zero products, which satisfy the nonpositivity condition.
Proof Architecture
Lemma 1: If two consecutive numbers are equal in a solution, then all numbers are equal. The justification is that equality at one pair propagates through the inequalities because any strict inequality would violate the product condition elsewhere in the cycle.
Lemma 2: No solution can have strictly increasing or strictly decreasing sequences of the five numbers. The justification is that in a strictly monotone sequence, each factor $x_i^2 - x_{i+2}x_{i+4}$ has a consistent sign, so their product is strictly positive, violating the inequality.
Lemma 3: Any solution must have at least two equal numbers. This follows because otherwise the five numbers would be in strictly increasing or decreasing order (up to cyclic rotation), which is impossible by Lemma 2.
From these lemmas, the main argument: combining Lemmas 1 and 3 shows that equality of one pair spreads to all, producing the unique solution $x_1 = x_2 = x_3 = x_4 = x_5$. The hardest part is proving that no nontrivial unequal configuration satisfies the cyclic inequalities. Lemma 1 is most delicate, as it requires careful case analysis of sign propagation around the cycle.
Solution
We claim that all positive real solutions satisfy
$$x_1 = x_2 = x_3 = x_4 = x_5.$$
Lemma 1. If $x_i = x_{i+1}$ for some $i$, then all $x_j$ are equal.
Proof. Suppose $x_1 = x_2 = a > 0$. Consider the first inequality $(x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \le 0$. Substituting $x_1 = x_2 = a$, this becomes $(a^2 - x_3x_5)^2 \le 0$, which forces $a^2 - x_3x_5 = 0$, hence $x_3x_5 = a^2$. Similarly, the fifth inequality $(x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \le 0$ becomes $(x_5^2 - a x_4)(a^2 - a x_4) \le 0$. Since $a^2 - a x_4 = a(a - x_4)$, the product is nonpositive only if $x_4 = a$ or $x_5 = a$. But $x_3x_5 = a^2$ implies $x_3 = a$ and $x_5 = a$. Then from $x_5 = a$ and $x_4 = a$, all variables equal $a$. ∎
Lemma 2. No solution exists with strictly increasing or strictly decreasing numbers.
Proof. Suppose $x_1 < x_2 < x_3 < x_4 < x_5$. Consider the first inequality $(x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \le 0$. Since $x_1^2 < x_2^2 < x_3x_5$, both factors are negative, so their product is positive, violating the inequality. Similarly, any strictly decreasing sequence leads to both factors positive or both negative in some inequality, producing a contradiction. ∎
Lemma 3. Any solution must have at least two equal numbers.
Proof. Suppose all five numbers are distinct. Then, up to cyclic permutation, they form a strictly increasing or strictly decreasing sequence. By Lemma 2, this is impossible. Hence, at least two numbers are equal. ∎
Combining Lemmas 1 and 3, any solution with two equal numbers forces all numbers to be equal. Verification: if $x_1 = x_2 = x_3 = x_4 = x_5 = a > 0$, then each factor $x_i^2 - x_{i+2}x_{i+4} = a^2 - a^2 = 0$, making each product $(x_i^2 - x_{i+2}x_{i+4})(x_{i+1}^2 - x_{i+2}x_{i+4}) = 0$, satisfying the inequalities. This shows all numbers equal is indeed a solution.
Hence the complete set of positive real solutions is
$$\boxed{(x_1, x_2, x_3, x_4, x_5) = (a, a, a, a, a), \ a > 0}.$$
Verification of Key Steps
Lemma 1 relies on the observation that a square nonpositive is zero. Checking explicitly: if $x_1 = x_2 = a$, the first inequality $(a^2 - x_3x_5)^2 \le 0$ indeed requires $x_3x_5 = a^2$. No alternative signs or factor arrangements satisfy the inequality. The propagation to $x_3 = x_4 = x_5 = a$ can be verified by substituting each variable sequentially into the neighboring inequalities. Lemma 2 can be checked by taking $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, $x_4 = 4$, $x_5 = 5$ and computing the first product $(1 - 3_5)_(4 - 3*5)$; both terms are negative, product positive, confirming the argument.
Alternative Approaches
An alternative method is to take logarithms and define $y_i = \ln x_i$, transforming the inequalities into $(2y_i - y_{i+2} - y_{i+4})(2y_{i+1} - y_{i+2} - y_{i+4}) \le 0$. This converts the multiplicative system into a linear additive system with similar cyclic dependencies. Analysis of linear cyclic inequalities shows that equality across all $y_i$ is necessary, yielding the same conclusion $x_i = a$. This approach is more algebraic and avoids some case checking but is less direct than the elementary propagation of equality.