# 2013 USAJMO Problems/Problem 6

## Problem 6

Find all real numbers satisfying

## 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