Problem

[bugzpodder]

find all pairs of real (a,b,c) that are solution to:
a/b + b/a + c/a + a/c + b/c + c/b = 6

a/(bc) + b/(ac) + c/(ab) =4

we have from equation 1:

(a-b)^2/(ab)+(b-c)^2/(bc)+(c-a)^2/(ca)=0

now from second equation we have:
(a^2+b^2+c^2)/(abc)=4

so either a,b,c are all positive or two of a,b,c must be negative. i attempted the case when a,b are negative

a>0, b>0, c>0 implies that a=b=c=3/4

but what happens when two of them are negative? i tried to play around with the equations but got no where.

For those of you who has seen the problem before, it is found at the problem site: http://www.theproblemsite.com/maml/

of course, i have NO interest in taking a solution and submit it as my own.

[sql_lall]

AMGM:

a/b +b/c + c/a + b/a + c/b + a/c >= 6sqrt((abcbca)/(bcaabc))
=6

=> a/b +b/c + c/a + b/a + c/b + a/c =6 only when a=b=c

=> 3(1/a) = 4, a =3/4 as you found.

However, i'm not sure if that application of AMGM still works with negative numbers.

[bugzpodder]

well you could try it. x=-1 y=-1

AM=-1
GM=1