Difference between revisions of "2013 USAMO Problems/Problem 4"
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− | == Solution | + | == Solution (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>. | ||
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This is proven easily. | This is proven easily. | ||
<cmath>\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}</cmath> by Cauchy. | <cmath>\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}</cmath> by Cauchy. | ||
+ | |||
Equality then holds when <math>a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1</math>. | Equality then holds when <math>a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1</math>. | ||
Now assume that <math>x = \min(x,y,z)</math>. Now note that, by the Lemma, | Now assume that <math>x = \min(x,y,z)</math>. Now note that, by the Lemma, | ||
− | <cmath>\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}</cmath>. So equality must hold. | + | <cmath>\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}</cmath>. So equality must hold in order for the condition in the problem statement to be met. |
So <math>(y-1)(z-1) = 1</math> and <math>(x-1)(yz) = 1</math>. If we let <math>z = c</math>, then we can easily compute that <math>y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}</math>. | So <math>(y-1)(z-1) = 1</math> and <math>(x-1)(yz) = 1</math>. If we let <math>z = c</math>, then we can easily compute that <math>y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}</math>. | ||
Now it remains to check that <math>x \le y, z</math>. | Now it remains to check that <math>x \le y, z</math>. | ||
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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: | ||
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 2 == | ||
+ | WLOG, assume that <math>x = \min(x,y,z)</math>. Let <math>a=\sqrt{x-1},</math> <math>b=\sqrt{y-1}</math> and <math>c=\sqrt{z-1}</math>. Then <math>x=a^2+1</math>, <math>y=b^2+1</math> and <math>z=c^2+1</math>. The equation becomes | ||
+ | <cmath>(a^2+1)+(a^2+1)(b^2+1)(c^2+1)=(a+b+c)^2.</cmath> | ||
+ | Rearranging the terms, we have | ||
+ | <cmath>(1+a^2)(bc-1)^2+[a(b+c)-1]^2=0.</cmath> | ||
+ | Therefore <math>bc=1</math> and <math>a(b+c)=1.</math> Express <math>a</math> and <math>b</math> in terms of <math>c</math>, we have <math>a=\frac{c}{c^2+1}</math> and <math>b=\frac{1}{c}.</math> Easy to check that <math>a</math> is the smallest among <math>a</math>, <math>b</math> and <math>c.</math> Then <math>x=\frac{c^4+3c^2+1}{(c^2+1)^2}</math>, <math>y=\frac{c^2+1}{c^2}</math> and <math>z=c^2+1.</math> | ||
+ | Let <math>c^2=t</math>, we have the solutions for <math>(x,y,z)</math> as follows: | ||
+ | <math>(\frac{t^2+3t+1}{(t+1)^2}, \frac{t+1}{t}, t+1)</math> and permutations for all <math>t>0.</math> | ||
+ | |||
+ | --J.Z. | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:43, 20 April 2023
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 in order for the condition in the problem statement to be met.
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
.
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.