Difference between revisions of "1969 IMO Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | Prove that for all real numbers <math>x_1, x_2, y_1, y_2, z_1, z_2</math>, with <math>x_1 > 0, x_2 | + | Prove that for all real numbers <math>x_1, x_2, y_1, y_2, z_1, z_2</math>, with <math>x_1 > 0, x_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0</math>, the inequality<cmath>\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}</cmath>is satisfied. Give necessary and sufficient conditions for equality. |
==Solution== | ==Solution== | ||
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and replace in [Equation 3] to get: | 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}} | + | <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 | + | 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}} | {{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}} |
Revision as of 17:51, 27 July 2024
Contents
Problem
Prove that for all real numbers , with
, the inequality
is 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 |