Difference between revisions of "1975 IMO Problems/Problem 1"
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Let <math>x_i, y_i</math> <math>(i=1,2,\cdots,n)</math> be real numbers such that <cmath>x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.</cmath> Prove that, if <math>z_1, z_2,\cdots, z_n</math> is any permutation of <math>y_1, y_2, \cdots, y_n,</math> then <cmath>\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2.</cmath> | Let <math>x_i, y_i</math> <math>(i=1,2,\cdots,n)</math> be real numbers such that <cmath>x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.</cmath> Prove that, if <math>z_1, z_2,\cdots, z_n</math> is any permutation of <math>y_1, y_2, \cdots, y_n,</math> then <cmath>\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2.</cmath> | ||
− | == | + | ==Solution1== |
We can rewrite the summation as | We can rewrite the summation as | ||
<cmath>\sum^n_{i=1} x_i^2 + \sum^n_{i=1} y_i^2 - \sum^n_{i=1}x_iy_i \le \sum^n_{i=1} x_i^2 + \sum^n_{i=1} z_i^2 - \sum^n_{i=1}x_iz_i.</cmath> | <cmath>\sum^n_{i=1} x_i^2 + \sum^n_{i=1} y_i^2 - \sum^n_{i=1}x_iy_i \le \sum^n_{i=1} x_i^2 + \sum^n_{i=1} z_i^2 - \sum^n_{i=1}x_iz_i.</cmath> | ||
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<cmath> </cmath> | <cmath> </cmath> | ||
~Imajinary | ~Imajinary | ||
+ | ==Notice== | ||
+ | It is only the most common way of rearrangement inequality after expanding and subtracting same terms.~bluesoul | ||
+ | |||
==See Also== | ==See Also== | ||
{{IMO box|year=1975|before=First Question|num-a=3}} | {{IMO box|year=1975|before=First Question|num-a=3}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 14:17, 19 December 2021
Contents
[hide]Problem
Let be real numbers such that Prove that, if is any permutation of then
Solution1
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
Notice
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 |