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.

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 $F.$
