2013 USAMO Problems/Problem 4

Find all real numbers $x,y,z\geq 1$ satisfying $\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.$

Solution (Cauchy or AM-GM)

The key Lemma is: $\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}$ for all $a,b \ge 1$ . Equality holds when $(a-1)(b-1)=1$ .

This is proven easily. $\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}$ by Cauchy. Equality then holds when $a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1$ .

Now assume that $x = \min(x,y,z)$ . Now note that, by the Lemma,

$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}$ . So equality must hold. So $(y-1)(z-1) = 1$ and $(x-1)(yz) = 1$ . If we let $z = c$ , then we can easily compute that $y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}$ . Now it remains to check that $x \le y, z$ .

But by easy computations, $x = \frac{c^2+c-1}{c^2} \le c = z \Longleftrightarrow (c^2-1)(c-1) \ge 0$ , which is obvious. Also $x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1$ , which is obvious, since $c \ge 1$ .

So all solutions are of the form $\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}$ , and symmetric (or cyclic) permutations for $c > 1$ .

Remark: An alternative proof of the key Lemma is the following: By AM-GM, $(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}$ $ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}$ . Now taking the square root of both sides gives the desired. Equality holds when $(a-1)(b-1) = 1$ . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2013 USAMO Problems/Problem 4

Solution (Cauchy or AM-GM)