Difference between revisions of "2013 USAMO Problems/Problem 4"
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− | == Solution 1 ( | + | == Solution 1 (Cauchy or AM-GM) == |
The key Lemma is: | The key Lemma is: | ||
<cmath>\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}</cmath> for all <math>a,b \ge 1</math>. Equality holds when <math>(a-1)(b-1)=1</math>. | <cmath>\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}</cmath> for all <math>a,b \ge 1</math>. Equality holds when <math>(a-1)(b-1)=1</math>. |
Revision as of 20:22, 11 May 2013
Find all real numbers satisfying
Solution 1 (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 symmetric (or cyclic) 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
.