Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 USAMO Problems/Problem 4"

(→‎Solution 1 (Cauchy or AM-GM))
m
 
(8 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
== Problem ==
 +
 
Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath>
 
Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath>
  
−
 
+
== Solution (Cauchy or AM-GM) ==
−
== 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>.
Line 8: Line 9:
 
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>.
Line 19: Line 21:
 
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 symmetric (or cyclic) permutations for <math>c > 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 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 13:36, 28 October 2024

Problem

Find all real numbers $x,y,z\geq 1$ satisfying \[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]

Solution (Cauchy or AM-GM)

The key Lemma is: \[\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}\] for all $a,b \ge 1$. Equality holds when $(a-1)(b-1)=1$.

This is proven easily. \[\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}\] by Cauchy.

Equality then holds when $a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1$.

Now assume that $x = \min(x,y,z)$. Now note that, by the Lemma,

\[\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}\]. So equality must hold in order for the condition in the problem statement to be met. So $(y-1)(z-1) = 1$ and $(x-1)(yz) = 1$. If we let $z = c$, then we can easily compute that $y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}$. Now it remains to check that $x \le y, z$.

But by easy computations, $x = \frac{c^2+c-1}{c^2} \le c = z \Longleftrightarrow (c^2-1)(c-1) \ge 0$, which is obvious. Also $x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1$, which is obvious, since $c \ge 1$.

So all solutions are of the form $\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}$, and all permutations for $c > 1$.

Remark: An alternative proof of the key Lemma is the following: By AM-GM, \[(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}\] \[ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}\]. Now taking the square root of both sides gives the desired. Equality holds when $(a-1)(b-1) = 1$.

Solution 2

WLOG, assume that $x = \min(x,y,z)$. Let $a=\sqrt{x-1},$ $b=\sqrt{y-1}$ and $c=\sqrt{z-1}$. Then $x=a^2+1$, $y=b^2+1$ and $z=c^2+1$. The equation becomes \[(a^2+1)+(a^2+1)(b^2+1)(c^2+1)=(a+b+c)^2.\] Rearranging the terms, we have \[(1+a^2)(bc-1)^2+[a(b+c)-1]^2=0.\] Therefore $bc=1$ and $a(b+c)=1.$ Express $a$ and $b$ in terms of $c$, we have $a=\frac{c}{c^2+1}$ and $b=\frac{1}{c}.$ Easy to check that $a$ is the smallest among $a$, $b$ and $c.$ Then $x=\frac{c^4+3c^2+1}{(c^2+1)^2}$, $y=\frac{c^2+1}{c^2}$ and $z=c^2+1.$ Let $c^2=t$, we have the solutions for $(x,y,z)$ as follows: $(\frac{t^2+3t+1}{(t+1)^2}, \frac{t+1}{t}, t+1)$ and permutations for all $t>0.$

--J.Z.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_4&oldid=230738"