Solve a+b+c=0 given a+(1/b)=b+(1/c)=c+(1/a)

[Perpendicular]

Given : a + (1/b) = b + (1/c) = c + (1/a)

prove a=b=c or a+b+c = 0

I know that a^3 + b^3 + c^3 = 3abc is sufficient proof to derive the above two relations asked for..but I cannot get the expression up to there.

I tried to see if a+b+c = 0 logically leads to the given equation, but no luck so far.

What am I missing ?

 

[dikmikkel]

Assume that the equation is true, and assume that first a solution is a=b=c try replacing a with b:
b+1/b = b+1/c=c+1/b
And c should also be equal to b. So replace again:
c+1/c = c+1/c = c+1/c
True true true. So indeed that is a solution. Do the same for a+b+c=0 <=> a = -b-c
and find that the equation is true

 

[SammyS]

Given : a + (1/b) = b + (1/c) = c + (1/a)

prove a=b=c or a+b+c = 0

I know that a^3 + b^3 + c^3 = 3abc is sufficient proof to derive the above two relations asked for..but I cannot get the expression up to there.

I tried to see if a+b+c = 0 logically leads to the given equation, but no luck so far.

What am I missing ?

You have two independent equations:

a + (1/b) = b + (1/c)​

and

a + (1/b) = c + (1/a)
​

Solve the second one for c, and plug that into the first one.

Do some algebraic manipulation. One solution should be a = b .

The other solution is more complicated. I assume it should lead to a+b+c=0

 

[tiny-tim]

hmm … the solution is symmetric, so i'd expect the proof to be symmetric also

if we subtract the three equations in pairs, i can get to abc = ±1, or a = b = c

and it might help to use that if p/q = r/s = t/u, then they all equal [Ap+Br+Ct]/[Aq+Bs+Cu]

 

[Perpendicular]

Putting a+b+c = 0 the equation is not becoming an identity...

 

[SammyS]

Putting a+b+c = 0 the equation is not becoming an identity...

What else have you tried?