How can this possibly be unique...

$(a+b+c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ca), \\ 4 = 4 + 2(ab+bc+ca), \\ ab + bc + ca = 0.$ Hence, $\begin{aligned} 0 &= (ab + bc + ca)^2 \\ &= a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c), \\ a^2b^2 + b^2c^2 + c^2a^2 &= -4abc. \end{aligned}$ And so $\begin{aligned} \frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a} &= \frac{a^2b^2 + b^2c^2 + c^2a^2}{abc} \\ &= \frac{-4abc}{abc} \\ &= \boxed{-4}. \end{aligned}$

I did slightly different.

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

$2^2 = 4 + 2(ab+bc+ca) \quad \Rightarrow ab+bc+ca = 0$

$\Rightarrow \dfrac {ab+bc+ca}{c} = 0 \quad \Rightarrow \dfrac {ab}{c} + b + a= 0$

$\Rightarrow \dfrac {ab}{c} = -(a+b)$

Similarly,

$\dfrac {bc}{a} = -(b + c) \quad$ and $\quad \dfrac {ca}{b} = -(c + a)$

$\Rightarrow \dfrac {ab}{c} + \dfrac {bc}{a} + \dfrac {ca}{b} = -2(a+b+c) = \boxed{-4}$

(a+b+c)^2=2^2; Solve above equation we get, ab+bc+ca=0; THEN again squaring on both sides of the result ab+bc+ca=0 Then solve we get, (ab) ^2 +(bc) ^2 +(ca)^2+2 abc[a+b+c]=0 Then divide on both by any we get, ab/c+ac/b+bc/a=-2(a+b+c)=-2 2=-4

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

Which becomes $a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2$

So $ab+bc+ac=0$

Separating them out:

$ab=-ac-bc \longrightarrow {\frac{ab}{c}=-a-b}$

$ac=-ab-bc \longrightarrow {\frac{ac}{b}=-a-c}$

$bc=-ab-ac \longrightarrow{\frac{bc}{a}=-b-c}$

Adding up the equations:

$\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}=-2a-2b-2c=-2(a+b+c)=\boxed{-4}$

(a+b+c) 2 = a 2 +b 2 + c 2 +2(ab+bc+ca) = 4 as given that implies ab+bc+ca = 0 and again squaring ab+bc+ca we get (ab) 2 +(bc) 2+(ca) 2 +2abc(a+b+c) that implies (ab) 2+(bc) 2+(ca) 2 =-4abc substituting in given (ab/c)+(bc/a)+(ca/b) gives the value -4 {* indicates power}

a+b+c=2

(a+b+c)²=4=a²+b²+c²+2(ab+bc+ac)

Which gives ab+bc+ac=0

Dividing by a, b & c gives

b+(bc/a)+c=0

a+c+(ac/b)=0

(ab/c)+b+a=0

Adding all 3 gives

(ab/c)+(bc/a)+(ac/b) = -2(a+b+c) = -4