2013 USAMO Problems/Problem 4

Revision as of 20:21, 11 May 2013 by Mathocean97 (talk | contribs)

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 1 (Standard Inequalities)

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 $(a-1)(b-1) = 1$.

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