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Only one, one, one (Posted on 2007-05-30) Difficulty: 3 of 5
It's obvious that x=y=z=1 is an all-positive, all-real, solution of

x + y + z = 3
y + z + x = 3
z + x + y = 3

Can you prove that it is the only such solution?

See The Solution Submitted by Federico Kereki    
Rating: 4.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 3 of 5 |

Given x>0, y>0, z>0 is is obvious that if any of x,y,z is greater than 1 then another must be less than 0 and vice versa.  Then problem can be divided into four cases depending on how many of x,y,z are greater than or less than 1.

- Case 1: two of x,y,z are greater than 1, and the third is less than 1.
Without loss of generality let x>1, y>1, 1>z.  Then x^2>x, y^3>y^2, z>z^3, which implies x^2+y^3+z > x+y^2+z^3.  But from the equations given, the inequality should be equal, therefore there are no solutions from this case

- Case 2: two of x,y,z are less than 1, and the third is greater than 1.
Without loss of generality let x>1, 1>y, 1>z.  Then x^3>x, y>y^2, z^2>z^3, which implies x^3+y+z^2 > x+y^2+z^3.  But from the equations given, the inequality should be equal, therefore there are no solutions from this case

- Case 3: one of x,y,z is less than 1, one is greater than 1, and one equals 1.
Without loss of generality let x=1.  Then this case can be divided into two subcases.
- - Subcase 1: x=1, y>1, 1>z
Then y^3>y and z>z^2, which implies 1+y^3+z>1+y>z^2.  But from the equations given, the inequality should be equal, therefore there are no solutions from this subcase.
- - Subcase 2: x=1, 1>y, z>1
Then y>y^3 and z^2>2, which implies 1+y>z^2>1+y^3+z.  But from the equations given, the inequality should be equal, therefore there are no solutions from this subcase.

- Case 4: x=y=z=1
This is obviously a solution, and since no solutions were found in any other case, this is the only solution.

Edited on May 31, 2007, 2:55 pm
  Posted by Brian Smith on 2007-05-31 14:53:35

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