Difference between revisions of "1975 IMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
We can expand and simplify the inequality a bit, and using the fact that <math>z</math> is a permutation of <math>y</math>, we can cancel some terms. | We can expand and simplify the inequality a bit, and using the fact that <math>z</math> is a permutation of <math>y</math>, we can cancel some terms. | ||
− | <cmath>\sum^n_{i=1}x_i^2 + \sum^n{i=1}y_i^2 - | + | <cmath>\sum^n_{i=1}x_i^2 + \sum^n{i=1}y_i^2 - 2\sum^n{i=1}x_iy_i \leq \sum^n_{i=1}x_i^2 + \sum^n{i=1}z_i^2 - 2\sum^n{i=1}x_iz_i</cmath> |
− | <cmath>sum^n{i=1}x_iy_i \geq sum^n{i=1}x_iz_i</cmath> | + | <cmath>\sum^n{i=1}x_iy_i \geq \sum^n{i=1}x_iz_i</cmath> |
Consider the pairing <math>x_1 \rightarrow y_1</math>, <math>x_2 \rightarrow y_2</math>, ... <math>x_n \rightarrow y_n</math>. By switching around some of the <math>y</math> values, we have obtained the pairing <math>x_1 \rightarrow z_1</math>, <math>x_2 \rightarrow z_2</math>, ... <math>x_n \rightarrow z_n</math>. Suppose that we switch around two <math>y</math>-values, <math>y_m</math> and <math>y_n</math>, such that <math>y_m > y_n</math>. If <math>x_m > x_n</math>, call this a type 1 move. Otherwise, call this a type 2 move. | Consider the pairing <math>x_1 \rightarrow y_1</math>, <math>x_2 \rightarrow y_2</math>, ... <math>x_n \rightarrow y_n</math>. By switching around some of the <math>y</math> values, we have obtained the pairing <math>x_1 \rightarrow z_1</math>, <math>x_2 \rightarrow z_2</math>, ... <math>x_n \rightarrow z_n</math>. Suppose that we switch around two <math>y</math>-values, <math>y_m</math> and <math>y_n</math>, such that <math>y_m > y_n</math>. If <math>x_m > x_n</math>, call this a type 1 move. Otherwise, call this a type 2 move. | ||
Revision as of 00:00, 10 December 2022
Contents
[hide]Problem
Let be real numbers such that Prove that, if is any permutation of then
Solution
We can expand and simplify the inequality a bit, and using the fact that is a permutation of , we can cancel some terms. Consider the pairing , , ... . By switching around some of the values, we have obtained the pairing , , ... . Suppose that we switch around two -values, and , such that . If , call this a type 1 move. Otherwise, call this a type 2 move.
Type 2 moves only increase the sum of the products of the pairs. The sum is increased by . This is equivalent to , which is clearly nonnegative since and .
We will now consider switching from the - pairing back to the - pairing. We will prove that from any pairing of and values, you can use just type 2 moves to navigate back to the pairing of and values, which will complete the proof.
Suppose that is the biggest -value that is not paired with its -value in the and pairing. Then, switch this -value with the -value currently paired with . This is obviously a type 2 move. Continue this process until you reach back to the - pairing. All moves are type 2 moves, so the proof is complete.
~mathboy100
Solution 2
We can rewrite the summation as Since , the above inequality is equivalent to We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of . Obviously, if or , the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of s that are not ordered in the form such that is at a maximum out of all possible permutations and is greater than the sum . This necessarily means that in the sum there exists two terms and such that and . Notice that which means if we make the terms and instead of the original and , we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality is proved, which is equivalent to what we wanted to prove. ~Imajinary
Remark
It is only the most common way of rearrangement inequality after expanding and subtracting same terms.~bluesoul
See Also
1975 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |