Difference between revisions of "1969 IMO Problems/Problem 6"
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− | <math>\sqrt{AB} \le \frac{A+B}{2}</math> | + | <math>\sqrt{AB} \le \frac{A+B}{2}</math> with equality at <math>A=B</math> |
<math>4AB \le (A+B)^2</math> | <math>4AB \le (A+B)^2</math> | ||
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<math>\frac{8}{2(A+B)} \le \frac{A+B}{AB}</math> | <math>\frac{8}{2(A+B)} \le \frac{A+B}{AB}</math> | ||
− | <math>\frac{8}{2(A+B)} \le \frac{1}{A}+\frac{1}{B}</math> | + | <math>\frac{8}{2(A+B)} \le \frac{1}{A}+\frac{1}{B}</math> [Equation 1] |
+ | <math>(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2</math> | ||
− | {{solution}} | + | <math>(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=(A+B)+x_1y_2+x_2y_1-2z_1z_2</math> |
+ | |||
+ | since <math>x_1y_1>z_1^2</math> and <math>x_2y_2>z_2^2</math>, and using the Rearrangement inequality | ||
+ | |||
+ | then <math>x_1y_1+x_2y_2-z_1z_1-z_2z_2 \le x_1y_2+x_2y_1-z_1z_2-z_1z_2</math> | ||
+ | |||
+ | <math>(A+B) \le x_1y_2+x_2y_1-2z_1z_2</math> | ||
+ | |||
+ | <math>2(A+B) \le x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2</math> | ||
+ | |||
+ | <math>2(A+B) \le (x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2</math> [Equation 2] | ||
+ | |||
+ | Therefore, we can can use [Equation 2] into [Equation 1] to get: | ||
+ | |||
+ | <math>\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \le \frac{1}{A}+\frac{1}{B}</math> | ||
+ | |||
+ | Then, from the values of <math>A</math> and <math>B</math> we get: | ||
+ | |||
+ | <math>\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}</math> | ||
+ | |||
+ | With equality at <math>x_1y_1 - z_1^2=x_2y_2 - z_2^2>0</math> and <math>x_1=x_2, y_1=y_2, z_1=z_2</math> | ||
+ | |||
+ | ~Tomas Diaz. orders@tomasdiaz.com | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | This solution is actually more difficult but I added it here for fun to see the generalized case as follows: | ||
+ | |||
+ | Prove that for all real numbers <math>a_i, b_i</math>, for <math>i=1,2,...,n</math> with <math>a_i > 0, b_i > 0</math> | ||
+ | |||
+ | and <math>\prod_{i=1}^{n-1}a_i-a_n^{n-1}>0, \prod_{i=1}^{n-1}b_i-b_n^{n-1}>0</math> the inequality | ||
+ | |||
+ | <cmath>\frac{2^n}{\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \leq \frac{1}{\prod_{i=1}^{n-1}a_i-a_n^{n-1}} + \frac{1}{\prod_{i=1}^{n-1}b_i-b_n^{n-1}}</cmath>is satisfied. | ||
+ | |||
+ | Let <math>A=\prod_{i=1}^{n-1}a_i-a_n^{n-1}</math> and <math>\prod_{i=1}^{n-1}b_i-b_n^{n-1}>0</math> | ||
+ | |||
+ | From AM-GM: | ||
+ | |||
+ | <math>\sqrt{AB} \le \frac{A+B}{2}</math> with equality at <math>A=B</math> | ||
+ | |||
+ | <math>4AB \le (A+B)^2</math> | ||
+ | |||
+ | <math>\frac{4}{A+B} \le \frac{A+B}{AB}</math> | ||
+ | |||
+ | <math>\frac{2^n}{2^{n-2}(A+B)} \le \frac{A+B}{AB}</math> | ||
+ | |||
+ | <math>\frac{2^n}{2^{n-2}(A+B)} \le \frac{1}{A}+\frac{1}{B}</math> [Equation 3] | ||
+ | |||
+ | Here's the difficult part where I'm skipping steps: | ||
+ | |||
+ | we prove that <math>2^{n-2}(A+B) \le \prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}</math> | ||
+ | |||
+ | and replace in [Equation 3] to get: | ||
+ | |||
+ | <math>\frac{2^n}{\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \le \frac{1}{A}+\frac{1}{B}</math> | ||
+ | |||
+ | and replace the values of <math>A</math> and <math>B</math> to get: | ||
+ | |||
+ | <math>\frac{2^n}{\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \leq \frac{1}{\prod_{i=1}^{n-1}a_i-a_n^{n-1}} + \frac{1}{\prod_{i=1}^{n-1}b_i-b_n^{n-1}}</math> | ||
+ | |||
+ | with equality at <math>a_i=b_i</math> for all <math>i=1,2,...,n</math> | ||
+ | |||
+ | Then set <math>n=3, a_1=x_1, a_2=y_1, a_3=z_1, b_1=x_2, b_2=y_2, b_3=z_3</math> and substitute in the generalized inequality to get: | ||
+ | |||
+ | <math>\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}</math> | ||
+ | |||
+ | with equality at <math>x_1=x_2, y_1=y_2, z_1=z_2</math> | ||
+ | |||
+ | ~Tomas Diaz. orders@tomasdiaz.com | ||
+ | |||
+ | {{alternate solutions}} | ||
== See Also == {{IMO box|year=1969|num-b=5|after=Last Question}} | == See Also == {{IMO box|year=1969|num-b=5|after=Last Question}} |
Latest revision as of 04:21, 19 November 2023
Contents
Problem
Prove that for all real numbers , with , the inequalityis satisfied. Give necessary and sufficient conditions for equality.
Solution
Let and
From AM-GM:
with equality at
[Equation 1]
since and , and using the Rearrangement inequality
then
[Equation 2]
Therefore, we can can use [Equation 2] into [Equation 1] to get:
Then, from the values of and we get:
With equality at and
~Tomas Diaz. orders@tomasdiaz.com
Solution 2
This solution is actually more difficult but I added it here for fun to see the generalized case as follows:
Prove that for all real numbers , for with
and the inequality
is satisfied.
Let and
From AM-GM:
with equality at
[Equation 3]
Here's the difficult part where I'm skipping steps:
we prove that
and replace in [Equation 3] to get:
and replace the values of and to get:
with equality at for all
Then set and substitute in the generalized inequality to get:
with equality at
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1969 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |