Prove that if $abc\ne0$ and $ab+bc+ac=0$ then $a+b+c\ne0$

Question

I tried to do proof by contradiction, but problem is how to get from $ab+bc+ac$ to $a+b+c$

Assuming $a+b+c=0$ my approachs:

Adding $ab+ac+bc=0$ and $a+b+c=0$ and try to factor
Deriving $$a^2+ab+ac=0\\ac+bc+c^2=0\\ab+b^2+bc=0$$ and trying to derive something but nothing useful.

Please just some hint will do the help to me

I believe that we should add the assumption that these are real numbers (otherwise the roots of $X^3-1$ are a counterexample). — Andrew Dudzik, Oct 10 '14 at 21:39

Andrew Dudzik · Answer 1 · 2014-10-11T01:08:17.663

4

If $ab+bc+ac=a+b+c=0$, then $a,b,c$ are the three roots of a cubic equation $X^3 - abc=0$. But any nonzero number has two non-real cube roots, so $abc=0$—assuming that at least two of $a,b,c$ are real.

edited Oct 11 '14 at 01:08

answered Oct 10 '14 at 21:43

Andrew Dudzik

30,074

score 2 · Accepted Answer · answered Oct 10 '14 at 21:04

2

You asked for a hint. Try thinking about $(a+b+c)(a+b+c)$

answered Oct 10 '14 at 21:04

paw88789

40,402

okay I get $a+b+c=0$ and $a^2+b^2+c^2=0$ how do I go about proving that it $\implies a=b=c=0$ ? – user182452 Oct 10 '14 at 21:17
@user182452 we have $x+y+z = 0$ with $x,y,z \geq 0$. What do we know about $x,y,z?$ – Ben Grossmann Oct 10 '14 at 21:19
of course duh, finally understood thank you all :) – user182452 Oct 10 '14 at 21:21

Prove that if $abc\ne0$ and $ab+bc+ac=0$ then $a+b+c\ne0$

2 Answers2