Difference between revisions of "2013 USAMO Problems/Problem 4"
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By AM-GM, <cmath>(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}</cmath> | By AM-GM, <cmath>(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}</cmath> | ||
<cmath>ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}</cmath>. Now taking the square root of both sides gives the desired. Equality holds when <math>(a-1)(b-1) = 1</math>. | <cmath>ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}</cmath>. Now taking the square root of both sides gives the desired. Equality holds when <math>(a-1)(b-1) = 1</math>. | ||
+ | {{MAA Notice}} |
Revision as of 16:40, 3 July 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
.
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