1971 IMO Problems/Problem 1


Prove that the following assertion is true for $n=3$ and $n=5$, and that it is false for every other natural number $n>2:$

If $a_1, a_2,\cdots, a_n$ are arbitrary real numbers, then $(a_1-a_2)(a_1-a_3)\cdots (a_1-a_n)+(a_2-a_1)(a_2-a_3)\cdots (a_2-a_n)+\cdots+(a_n-a_1)(a_n-a_2)\cdots (a_n-a_{n-1})\ge 0.$


Take $a_1 < 0$, and the remaining $a_i = 0$. Then $E_n = a_1(n-1) < 0$ for $n$ even, so the proposition is false for even $n$.

Suppose $n \ge 7$ and odd. Take any $c > a > b$, and let $a_1 = a$, $a_2 = a_3 = a_4= b$, and $a_5 = a_6 = ... = a_n = c$. Then $E_n = (a - b)^3 (a - c)^{n-4} < 0$. So the proposition is false for odd $n \ge 7$.

Assume $a_1 \ge a_2 \ge a_3$. Then in $E_3$ the sum of the first two terms is non-negative, because $a_1 - a_3 \ge a_2 - a3$. The last term is also non-negative. Hence $E_3 \ge 0$, and the proposition is true for $n = 3$.

It remains to prove $S_5$. Suppose $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5$. Then the sum of the first two terms in $E_5$ is $(a_1 - a_2){(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)} \ge 0$. The third term is non-negative (the first two factors are non-positive and the last two non-negative). The sum of the last two terms is: $(a_4 - a_5){(a_1 - a_5)(a_2 - a_5)(a_3 - a_5) - (a_1 - a_4)(a_2 - a_4)(a_3 - a_4)} \ge 0$. Hence $E_5 \ge 0$.

This solution was posted and copyrighted by e.lopes. The original thread can be fond here: [1]

See Also

1971 IMO (Problems) • Resources
Preceded by
First Question
1 2 3 4 5 6 Followed by
Problem 2
All IMO Problems and Solutions
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