Difference between revisions of "1971 IMO Problems/Problem 1"
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==Solution== | ==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 fond here: [https://aops.com/community/p366761] | ||
==See Also== | ==See Also== | ||
− | {{IMO box|year= | + | {{IMO box|year=1971|before=First Question|num-a=2}} |
Latest revision as of 13:54, 29 January 2021
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 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 |