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

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--Thinking Process by suli | --Thinking Process by suli | ||

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+ | == 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}} |

## Revision as of 10:55, 16 May 2018

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 .

## 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.