Difference between revisions of "2013 USAJMO Problems/Problem 6"
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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>. | 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>. | ||
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| + | Suppose x = y = z. Then <math>\sqrt{x + x^3} = 3\sqrt{x-1}</math>, so <math>x + x^3 = 9x - 9</math>. It is easily verified that <math>x^3 - 8x + 9 = 0</math> has no solution in positive numbers greater than 1. Thus, <math>\sqrt{x + xyz} \ge \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math> for x = y = z. We suspect if the inequality always holds. | ||
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| + | Let x = 1. Then we have <math>\sqrt{1 + yz} \ge \sqrt{y-1} + \sqrt{z-1}</math>, which simplifies to <cmath>1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}</cmath> and hence <cmath>yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}</cmath> Let us try a few examples: if y = z = 2, we have <math>3 > 2</math>; if y = z, we have <math>y^2 - 2y + 3 \ge 2(y-1)</math>, which reduces to <math>y^2 - 4y + 5 \ge 0</math>. 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 <math>u = \sqrt{(y-1)(z-1)}</math>! Thus, <cmath>u^2 - 2u + 2 = (u-1)^2 + 1 > 0</cmath>, and the claim holds for x = 1. | ||
Revision as of 11:13, 14 April 2014
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 ![\[1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}\]](http://latex.artofproblemsolving.com/1/5/e/15ea895126a2309ec5d0795354dac079c7cc25f3.png)
and hence ![\[yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}\]](http://latex.artofproblemsolving.com/3/9/e/39eb443cc2a760eb03d2d95bfabdb1123da6759f.png)
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, ![\[u^2 - 2u + 2 = (u-1)^2 + 1 > 0\]](http://latex.artofproblemsolving.com/3/4/b/34bc0d41229e722d449cac8de681088b1eb6691d.png)
, and the claim holds for x = 1.
