Difference between revisions of "2018 USAMO Problems/Problem 1"
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The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete. | The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete. | ||
+ | |||
+ | - It should actually be 4(a)(a+b+c) + 4bc which results in a wrong inequality by AM-GM | ||
+ | |||
+ | Hence, the solution is wrong. | ||
==Solution 2== | ==Solution 2== |
Revision as of 01:20, 11 January 2025
Contents
Problem 1
Let be positive real numbers such that
. Prove that
Solution
WLOG let . Add
to both sides of the inequality and factor to get:
The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.
- It should actually be 4(a)(a+b+c) + 4bc which results in a wrong inequality by AM-GM
Hence, the solution is wrong.
Solution 2
https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg
-srisainandan6
Solution 3
Similarly to Solution 2, we will prove homogeneity but we will use that to solve the problem differently. Let . Note that
, thus proving homogeneity.
WLOG, we can scale down all variables such that the lowest one is . WLOG, let this be
.
We now have
, and we want to prove
Adding
to both sides and subtracting
gives us
, or
. Let
. Now, we have
By the trivial inequality, this is always true. Since all these steps are reversible, the proof is complete.
~SigmaPiE
2018 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |