Difference between revisions of "1969 IMO Problems/Problem 6"
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<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> | <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> | <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> |
Revision as of 03:21, 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 |