New-ton!

We can solve this by using newton's sum rule,

Let,

$S_1 = a + b + c$

$S_2 = a^2 + b^2 + c^2$

$S_3 = a^3 + b^3 + c^3$

$S_4 = a^4 + b^4 + c^4$

$S_5 = a^5 + b^5 + c^5$

$P_2 = ab + bc + ac$

$P_3 = abc$

Now,

According to newton's sum,

$S_2 = S_1^2 - 2P_2 \Rightarrow P_2 = -4$

$S_3 = S_1 S_2 - P_2 S_1 + 3 P_3 \Rightarrow P_3 = - 4$

Now,

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ ab + bc + ac }{ abc } = \frac{ P_2 }{ P_3 } = \frac{-4}{-4}$

$= \boxed{1}$

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$

$\Rightarrow \boxed{ab + bc + ca = -4}$

$a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$

$\Rightarrow \boxed{abc = -4}$

$\Rightarrow \dfrac{ab + bc + ca}{abc} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \boxed{1}$

A longer proof, that ignores newton sums.

Write $a,b,c$ as the roots of a monic polinomyal of degree $3$

$(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab + bc + ca)x - abc$

Let's try to guess the coefficients $\alpha_i$ of the $x^i$ in the polinomial from our assumptions.

$\alpha_2$ is clearly -1.

For $\alpha_1$ use

$(a+b+c)^2 = (a^2 + b^2 + c^2) + 2\alpha_1$

and obtain $\alpha_1 = - 4$ .

For $\alpha_0$ note that

$(a+b+c)(a^2 + b^2 + c^2) =$

$= a^3 + b^3 + c^3 + (a^2b + a^2c + b^2a +b^2c+ c^2a+ c^2b)$

hence

$(a^2b + a^2c + b^2a+b^2c+ c^2a+ c^2b) = 8$ .

Then use

$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2a+b^2c+ c^2a+ c^2b) + 6abc$

and get

$\alpha_0 = 4$ .

Now the polinomyal is

$x^3 - x^2 - 4x + 4 = (x^2 -4)(x-1) = (x-2)(x-1)(x+2)$

and its roots are $-2, \ \ 1, \ \ 2$

so the sum of their reciprocals is $1$ .

there should be c^2 Instead of c^3