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.

$p(x)=(x-a)(x-b)(x-c)=x^3-ux^2+vx-w$ has roots $a,b,c$ with sum $a+b+c=u$.
$(a+b)(b+c)(c+a)=0 \iff (u-c)(u-a)(u-b)=0 \iff p(u) = 0\,$ thus:
$$
\require{cancel}
\cancel{u^3}-\cancel{u\,u^2} + v\,u-w=0 \;\;\iff\;\; \frac{v}{w}=\frac{1}{u} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}
$$
