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
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