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.

Another approach, as you requested.
I write your question as  (L)$\iff $(R). Since (L) is clearly
valid if $a+b=0,$ it suffices to prove (L) implies (R).
I now assume that (L) is valid. By scaling, we can assume that $c=1.$
From
$$\frac{1}{a} +\frac{1}{b} +1=\frac{1}{a+b+1},$$
we get
$$\frac{a+b}{ab}=\frac{1}{a+b+1}-1=-\frac{a+b}{a+b+1}\tag1$$
We're done if $a+b=0$ so cancel $a+b$ in (1) to obtain
$-ab=a+b+1.$ Thus, $(a+1)(1+b)=0,$ proving (R).
(Scaling uses that the right side of (R) is zero.)
