# 2013 USAMO Problems/Problem 4

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.

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