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.

not an answer :
If we assume : $$a+b+c=abc$$
Then we obtain :
$$ab+bc+ca=1$$
Now substitute :
$$\frac{1}{a}=u$$
$$\frac{1}{b}=v$$
$$\frac{1}{c}=w$$
We obtain:
$$\frac{1}{uv}+\frac{1}{vw}+\frac{1}{wu}=1$$
Or :
$$uvw=u+v+w$$
So I recall the tangent formula :
$$\tan(a)+\tan(b)+\tan(c)=\tan(a)\tan(b)\tan(c)$$
So there is  a geometrical interpretation behind it .
