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.
 A: $p(x)=(x-a)(x-b)(x-c)=x^3-ux^2+vx-w$ 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:
$$
\require{cancel}
\cancel{u^3}-\cancel{u\,u^2} + v\,u-w=0 \;\;\iff\;\; \frac{v}{w}=\frac{1}{u} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}
$$
A: 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 :
$$(a+b)(b+c)(c+a)=0$$
A: 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 .
A: You could define:
$$F(a,b,c) = (a+b+c)(ab+bc+ca) - abc$$
and check that $F(a, -a,c) = F(a,b,-b) = F(-c,b,c) = 0$ easily. This means that $F$ as a polynomial must contain the linear factors $a+b, b+c$ and $c+a.$ Then, you argue that $F$ must be divisible by all of them at the same time, which generates a cubic polynomial where $F$ itself is cubic. Therefore,
$$F(a,b,c) = (a+b)(b+c)(c+a)$$
by matching a leading coefficient.
But this argument, while seems clever, requires unnecessary sophisticated methods from Number Theory/Algebra than just fully expanding $F.$
A: 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.)
