Difference between revisions of "2013 USAMO Problems/Problem 4"
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Also <math>x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1</math>, which is obvious, since <math>c \ge 1</math>. | Also <math>x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1</math>, which is obvious, since <math>c \ge 1</math>. | ||
− | So all solutions are of the form <math>\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}</math>, and | + | So all solutions are of the form <math>\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}</math>, and all permutations for <math>c > 1</math>. |
'''Remark:''' An alternative proof of the key Lemma is the following: | '''Remark:''' An alternative proof of the key Lemma is the following: |
Revision as of 11:20, 19 April 2014
Find all real numbers satisfying
Solution (Cauchy or AM-GM)
The key Lemma is:
for all
. Equality holds when
.
This is proven easily.
by Cauchy.
Equality then holds when
.
Now assume that . Now note that, by the Lemma,
. So equality must hold.
So
and
. If we let
, then we can easily compute that
.
Now it remains to check that
.
But by easy computations, , which is obvious.
Also
, which is obvious, since
.
So all solutions are of the form , and all permutations for
.
Remark: An alternative proof of the key Lemma is the following:
By AM-GM,
. Now taking the square root of both sides gives the desired. Equality holds when
.
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