All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math > Calculus
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 solution | Comment 1 of 5
Let's assume there is a solution which can be written as x = 1+d_x, y=1+d_y, z=1+d_z. The first equation, after some tinkering, then becomes d_x + 2 d_y + d_y^2 + 3d_z + 3 d_z^2 + d_z^3 = 0. The other equations give a similar result. Summing all three of these results gives:
d_x^3 + 4d_x^2+6d_x
+ d_y^3 + 4d_y^2+6d_y
+d_z^3 + 4d_z^2 + 6d_z = 0

The first part (the part with d_x) can be factored as d_x(d_x^2+4d_x+6). The only zero of this is d_x=0. The same goes for the other parts. The quadratic part will always be strictly positive, so the only solution of the equations is d_x=d_y=d_z=0, which results in x=y=z=1.

  Posted by Robby Goetschalckx on 2007-05-31 03:16:07
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information