<s>[INST] Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as  $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\iff (ab+bc+ca)(a+b+c)=abc \iff (a+b)(b+c)(c+a)=0  $
I'm curious to prove this some other ways. I tried to use function and inequality but still no progress.
 [/INST] We first start with a little rearrangement that is :
$$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} $$
implies :
$$ (ab+bc+ac)(a+b+c)-abc=0$$
Now note try setting $a=0$ we see then
$$bc(b+c)=0$$
So In order to make this vanish set ,
$$b=-c$$
But also observe that , this belongs to the solution set even if
$$a\neq 0$$
that is :
$$(a(b)+b(-b)+a(-b))(a+b-b)-ab(-b)=0$$
Hence by symmetry the other roots must be
$$a=-b , c= -a$$
Therefore we have that :
$$(a+b)(b+c)(c+a)=0$$
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