Difference between revisions of "2013 USAJMO Problems/Problem 6"
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Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath> | Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath> | ||
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==Solution with Thought Process== | ==Solution with Thought Process== | ||
Without loss of generality, let <math>1 \le 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>1 \le x \le y \le z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>. | ||
Revision as of 12:20, 20 April 2014
Problem 6
Find all real numbers 
satisfying ![\[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]](http://latex.artofproblemsolving.com/2/6/e/26edbd87f49eb68ec9b662b602db712e5151dc9c.png)
Solution
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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.
If x > 1, we see the 
will provide a huge obstacle when squaring. But, using the identity 
:
![\[x + xyz \ge x - 1 + y - 1 + z - 1 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(y-1)}\]](http://latex.artofproblemsolving.com/9/c/a/9ca4391efb211202afb93d416503511aee16aeca.png)
which leads to
![\[xyz \ge y + z - 3 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(z-1)}\]](http://latex.artofproblemsolving.com/2/0/3/203dc435bb009fd94b759d273476a2c6b3a0b1da.png)
Again, we experiment. If x = 2, y = 3, and z = 3, then 
.
Now, we see the finish: setting 
gives 
. 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:
![\[u^2(yz) - u(2\sqrt{y-1} + 2\sqrt{z-1}) + (3 + yz - y - z - 2\sqrt{(y-1)(z-1)}) \ge 0\]](http://latex.artofproblemsolving.com/b/b/a/bba13029b17812c8252f321e828e1905af983f89.png)
Because the coefficient of 
is positive, all we need to do is to verify that the discriminant is nonpositive:
![\[b^2 - 4ac = 4(y-1) + 4(z-1) - 8\sqrt{(y-1)(z-1)} - yz(12 + 4yz - 4y - 4z - 8\sqrt{(y-1)(z-1)})\]](http://latex.artofproblemsolving.com/9/f/d/9fdfdde4a15f73872fbe1f6c1f1fdedcf88fb796.png)
Let us try a few examples. If y = z, then the discriminant D = 
.
We are almost done, but we need to find the correct argument. (How frustrating!) Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis.
--Thinking Process by suli
