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2013 USAJMO Problems/Problem 6

Revision as of 10:42, 14 April 2014 by Suli (talk | contribs) (→‎Solution with Thought Process)

Solution with Thought Process

Without loss of generality, let $1 \le x \le y \le z$. Then $\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}$.

Suppose x = y = z. Then $\sqrt{x + x^3} = 3\sqrt{x-1}$, so $x + x^3 = 9x - 9$. It is easily verified that $x^3 - 8x + 9 = 0$ has no solution in positive numbers greater than 1. Thus, $\sqrt{x + xyz} \ge \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}$ for x = y = z. We suspect if the inequality always holds.

Let x = 1. Then we have $\sqrt{1 + yz} \ge \sqrt{y-1} + \sqrt{z-1}$, which simplifies to \[1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}\] and hence \[yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}\] Let us try a few examples: if y = z = 2, we have $3 > 2$; if y = z, we have $y^2 - 2y + 3 \ge 2(y-1)$, which reduces to $y^2 - 4y + 5 \ge 0$. 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 $u = \sqrt{(y-1)(z-1)}$! Thus, \[u^2 - 2u + 2 = (u-1)^2 + 1 > 0\] and the claim holds for x = 1.

If x > 1, we see the $\sqrt{x - 1}$ will provide a huge obstacle when squaring. But, using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz$: \[x + xyz \ge x - 1 + y - 1 + z - 1 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(y-1)}\] which leads to \[xyz \ge y + z - 3 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(z-1)}\] Again, we experiment. If x = 2, y = 3, and z = 3, then $18 > 7 + 4\sqrt{6}$.

Now, we see the finish: setting $u = \sqrt{x-1}$ gives $x = u^2 + 1$. 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: \[u^2(yz) - u(2\sqrt{y-1} + 2\sqrt{z-1}) + (3 + yz - y - z - 2\sqrt{(y-1)(z-1)}) \ge 0\]

Because u > 0, all we need to do is to verify that the discriminant is nonpositive: \[b^2 - 4ac = 4(y-1) + 4(z-1) - 8\sqrt{(y-1)(z-1)} - yz(12 - 4yz + 4y + 4z + 8\sqrt{(y-1)(z-1)})\] <prove that discriminant is negative> Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are $\boxed{no real solutions}$ to the original equation in the hypothesis. --Thinking Process by suli

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