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

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== Problem ==
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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>
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== Solution  (Cauchy or AM-GM) ==
 
== Solution  (Cauchy or AM-GM) ==
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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.
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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,
  
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<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.
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<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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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>.
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== Solution 2  ==
 
== 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
 
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

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.

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