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

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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 1 (Standard Inequalities) == | ||

+ | 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>. | ||

+ | |||

+ | 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. | ||

+ | 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, | ||

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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. | ||

+ | 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>. | ||

+ | |||

+ | But by easy computations, <math>x = \frac{c^2+c-1}{c^2} \le c = z \Longleftrightarrow (c^2-1)(c-1) \ge 0</math>, which is obvious. | ||

+ | 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>. | ||

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+ | 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>. | ||

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+ | '''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> | ||

+ | <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>. |

## Revision as of 20:21, 11 May 2013

Find all real numbers satisfying

## Solution 1 (Standard Inequalities)

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 .