Difference between revisions of "1975 IMO Problems/Problem 1"
Durianaops (talk | contribs) (Created page with "==Problem== 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...") |
|||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
− | {{ | + | 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> | |
+ | Since <math>\sum^n_{i=1} y_i^2 = \sum^n_{i=1} z_i^2</math>, the above inequality is equivalent to | ||
+ | <cmath>\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i.</cmath> | ||
+ | We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of <math>\sum^n_{i=1}x_iz_i</math>. Obviously, if <math>x_1 = x_2 = \ldots = x_n</math> or <math>y_1 = y_2 = \ldots = y_n</math>, the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of <math>z_i</math>s that are not ordered in the form <math>z_1 \ge z_2 \ge \ldots \ge z_n</math> such that <math>\sum^n_{i=1}x_iz_i</math> is at a maximum out of all possible permutations and is greater than the sum <math>\sum^n_{i=1}x_iy_i</math>. This necessarily means that in the sum <math>\sum^n_{i=1}x_iz_i</math> there exists two terms <math>x_pz_m</math> and <math>x_qz_n</math> such that <math>x_p > x_q</math> and <math>z_m < z_n</math>. Notice that | ||
+ | <cmath>x_pz_n + x_qz_m - (x_pz_m + x_qz_n) = (x_p-x_q)(z_n-z_m) > 0,</cmath> | ||
+ | which means if we make the terms <math>x_pz_n</math> and <math>x_qz_m</math> instead of the original <math>x_pz_m</math> and <math>x_qz_n</math>, we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality | ||
+ | <cmath>\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i</cmath> | ||
+ | is proved, which is equivalent to what we wanted to prove. | ||
+ | <cmath> </cmath> | ||
+ | ~Imajinary | ||
==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]] |
Latest revision as of 23:13, 6 November 2020
Problem
Let be real numbers such that Prove that, if is any permutation of then
Solution
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
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 |