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>. | ||
− | ==Solution with Thought Process== | + | ==(Incorrect) Solution with Thought Process== |
Without loss of generality, let <math>1 \le x \le y \le z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>. | Without loss of generality, let <math>1 \le x \le y \le z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>. | ||
Revision as of 10:06, 22 March 2020
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
.
(Incorrect) Solution with Thought Process
Without loss of generality, let . Then
.
Suppose x = y = z. Then , so
. It is easily verified that
has no solution in positive numbers greater than 1. Thus,
for x = y = z. We suspect if the inequality always holds.
Let x = 1. Then we have , which simplifies to
and hence
Let us try a few examples: if y = z = 2, we have
; if y = z, we have
, which reduces to
. The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let
! Thus,
and the claim holds for x = 1.
If x > 1, we see the will provide a huge obstacle when squaring. But, using the identity
:
which leads to
Again, we experiment. If x = 2, y = 3, and z = 3, then
.
Now, we see the finish: setting gives
. We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations:
Because the coefficient of is positive, all we need to do is to verify that the discriminant is nonpositive:
Let us try a few examples. If y = z, then the discriminant D = .
We are almost done, but we need to find the correct argument. (How frustrating!) Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis.
--Thinking Process by suli
Solution 2
WLOG, assume that . Let
and
. Then
,
and
. The equation becomes
Rearranging the terms, we have
Therefore
and
Express
and
in terms of
, we have
and
Easy to check that
is the smallest among
,
and
Then
,
and
Let
, we have the solutions for
as follows:
and permutations for all
--J.Z.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.