Solve AM GM HM Inequality for a+b+c=0

[erisedk]

Homework Statement

If a+b+c=0 then ( (b-c)/a + (c-a)/b + (a-b)/c )( a/(b-c) + b/(c-a) + c/(a-b) ) is equal to:

Ans: 9

Homework Equations

AM>=GM>=HM
Equality holds when all numbers are equal.

The Attempt at a Solution

I tried using AM>=GM.
( (b-c)/a + (c-a)/b + (a-b)/c + a/(b-c) + b/(c-a) + c/(a-b) )/6 >= 1

The AM side doesn't seem to simplify. All the numbers on the GM side cancel leaving me with one. I know I need to convert it into some form of a+b+c=0 but can't figure out how.

 

[Ray Vickson]

Homework Statement

If a+b+c=0 then ( (b-c)/a + (c-a)/b + (a-b)/c )( a/(b-c) + b/(c-a) + c/(a-b) ) is equal to:

Ans: 9

Homework Equations

AM>=GM>=HM
Equality holds when all numbers are equal.

The Attempt at a Solution

I tried using AM>=GM.
( (b-c)/a + (c-a)/b + (a-b)/c + a/(b-c) + b/(c-a) + c/(a-b) )/6 >= 1

The AM side doesn't seem to simplify. All the numbers on the GM side cancel leaving me with one. I know I need to convert it into some form of a+b+c=0 but can't figure out how.

The AM/GM inequality is irrelevant in this problem, and does not even apply. The reason is that if a+b+c=0 and at least one of a,b,c is nonzero, then at least one of them is positive and at least one is negative. The AM/GM inequality applies only if all the cited numbers are of the same sign.

Anyway, put c = -a-b and grind it through (and yes, indeed, it is lengthy!).

 

[erisedk]

Oh ok!