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Difference between revisions of "1971 IMO Problems/Problem 1"

m (→‎Solution)
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If <math>a_1, a_2,\cdots, a_n</math> are arbitrary real numbers, then <math>(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.</math>
 
If <math>a_1, a_2,\cdots, a_n</math> are arbitrary real numbers, then <math>(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.</math>
 +
  
 
==Solution==
 
==Solution==
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Assume <math>a_1 \ge a_2 \ge a_3</math>.
 
Assume <math>a_1 \ge a_2 \ge a_3</math>.
−
Then in <math>E_3</math> the sum of the first two terms is non-negative, because <math>a_1 - a_3 \ge a_2 - a3</math>.
+
Then in <math>E_3</math> the sum of the first two terms is non-negative, because <math>a_1 - a_3 \ge a_2 - a_3</math>.
 
The last term is also non-negative.
 
The last term is also non-negative.
 
Hence <math>E_3 \ge 0</math>, and the proposition is true for <math>n = 3</math>.
 
Hence <math>E_3 \ge 0</math>, and the proposition is true for <math>n = 3</math>.
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Suppose <math>a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5</math>.
 
Suppose <math>a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5</math>.
 
Then the sum of the first two terms in <math>E_5</math> is
 
Then the sum of the first two terms in <math>E_5</math> is
−
<math>(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</math>.
+
<math>(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</math>.
 
The third term is non-negative (the first two factors are non-positive and the last two non-negative).
 
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:
 
The sum of the last two terms is:
−
<math>(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</math>.
+
<math>(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</math>.
 
Hence <math>E_5 \ge 0</math>.
 
Hence <math>E_5 \ge 0</math>.
  
 
This solution was posted and copyrighted by e.lopes. The original thread can be found here: [https://aops.com/community/p366761]
 
This solution was posted and copyrighted by e.lopes. The original thread can be found here: [https://aops.com/community/p366761]
 +
 +
 +
==Remarks (added by pf02, December 2024)==
 +
 +
1. As a public service, I fixed a few typos in the solution above.
 +
 +
2. To make the solution a little more complete, let us note that
 +
the assumptions <math>a_1 \ge a_2 \ge a_3</math> in case <math>n = 3</math> and
 +
<math>a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5</math> in case <math>n = 5</math> are perfectly
 +
legitimate.  A different ordering of these numbers could be reduced
 +
to this case by a simple change of notation: we would substitute
 +
<math>a_i</math> by <math>b_j</math> with the indexes for the <math>b</math>'s chosen in such a way
 +
that the inequalities above are true for the <math>b</math>'s.
 +
 +
3. Also, the inequality
 +
<math>(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</math>
 +
is true because <math>a_1 - a_2 \le 0</math>, and
 +
<math>(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5) \le 0</math>.
 +
To see this latter inequality, just notice that <math>a_1 - a_3 \le a_2 - a_3</math>,
 +
and similarly for the other pairs of factors.  The difference of the products
 +
is <math>\le 0</math> as desired.
 +
 +
4. By looking at the proof above, we can also see that for <math>n = 3</math>
 +
we have equality if an only if <math>a_1 = a_2 = a_3</math>.  For <math>n = 5</math>, we
 +
have equality if and only if <math>a_1 = a_2</math> and <math>a_3 = a_4 = a_5</math>,
 +
or <math>a_1 = a_2 = a_3</math> and <math>a_4 = a_5</math> (still assuming that
 +
<math>a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5</math>).
 +
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1971|before=First Question|num-a=2}}
 
{{IMO box|year=1971|before=First Question|num-a=2}}

Revision as of 18:01, 14 December 2024

Problem

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 $(a_1-a_2)(a_1-a_3)\cdots (a_1-a_n)+(a_2-a_1)(a_2-a_3)\cdots (a_2-a_n)+\cdots+(a_n-a_1)(a_n-a_2)\cdots (a_n-a_{n-1})\ge 0.$


Solution

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 $a_1 - a_3 \ge a_2 - a_3$. 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 found here: [1]


Remarks (added by pf02, December 2024)

1. As a public service, I fixed a few typos in the solution above.

2. To make the solution a little more complete, let us note that the assumptions $a_1 \ge a_2 \ge a_3$ in case $n = 3$ and $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5$ in case $n = 5$ are perfectly legitimate. A different ordering of these numbers could be reduced to this case by a simple change of notation: we would substitute $a_i$ by $b_j$ with the indexes for the $b$'s chosen in such a way that the inequalities above are true for the $b$'s.

3. Also, the inequality $(a_1 - a_2)[(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)] \ge 0$ is true because $a_1 - a_2 \le 0$, and $(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5) \le 0$. To see this latter inequality, just notice that $a_1 - a_3 \le a_2 - a_3$, and similarly for the other pairs of factors. The difference of the products is $\le 0$ as desired.

4. By looking at the proof above, we can also see that for $n = 3$ we have equality if an only if $a_1 = a_2 = a_3$. For $n = 5$, we have equality if and only if $a_1 = a_2$ and $a_3 = a_4 = a_5$, or $a_1 = a_2 = a_3$ and $a_4 = a_5$ (still assuming that $a_1 \ge a_2 \ge a_3 \ge a_4 \ge a_5$).


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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