Difference between revisions of "1971 IMO Problems/Problem 1"
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| + | ==Problem== | ||
Prove that the following assertion is true for <math>n=3</math> and <math>n=5</math>, and that it is false for every other natural number <math>n>2:</math> | Prove that the following assertion is true for <math>n=3</math> and <math>n=5</math>, and that it is false for every other natural number <math>n>2:</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> | 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== | ||
| + | Take <math>a_1 < 0</math>, and the remaining <math>a_i = 0</math>. Then <math>E_n = a_1(n-1) < 0</math> for <math>n</math> even, | ||
| + | so the proposition is false for even <math>n</math>. | ||
| + | |||
| + | Suppose <math>n \ge 7</math> and odd. Take any <math>c > a > b</math>, and let <math>a_1 = a</math>, <math>a_2 = a_3 = a_4= b</math>, | ||
| + | and <math>a_5 = a_6 = ... = a_n = c</math>. | ||
| + | Then <math>E_n = (a - b)^3 (a - c)^{n-4} < 0</math>. | ||
| + | So the proposition is false for odd <math>n \ge 7</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>. | ||
| + | The last term is also non-negative. | ||
| + | Hence <math>E_3 \ge 0</math>, and the proposition is true for <math>n = 3</math>. | ||
| + | |||
| + | It remains to prove <math>S_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 | ||
| + | <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 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>. | ||
| + | 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] | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{IMO box|year=1971|before=First Question|num-a=2}} | ||
Revision as of 00:16, 19 July 2024
Problem
Prove that the following assertion is true for 
and 
, and that it is false for every other natural number 
If 
are arbitrary real numbers, then 
Solution
Take 
, and the remaining 
. Then 
for 
even,
so the proposition is false for even .
Suppose 
and odd. Take any 
, and let 
, 
,
and 
.
Then 
.
So the proposition is false for odd .
Assume 
.
Then in 
the sum of the first two terms is non-negative, because 
.
The last term is also non-negative.
Hence 
, and the proposition is true for 
.
It remains to prove 
.
Suppose 
.
Then the sum of the first two terms in 
is
.
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:
.
Hence 
.
This solution was posted and copyrighted by e.lopes. The original thread can be found 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 | ||
