Difference between revisions of "2013 USAJMO Problems/Problem 6"

Revision as of 09:56, 14 April 2014

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

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	==Solution==		==Solution==
−	Without loss of generality, let <math>x \ge y \ge z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>.	+	Without loss of generality, let <math>x \le y \le z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>.