Can You Handle This One

Wow. This one was a real workout.

$\begin{cases} f(a,b,c) = 2a^2+2b^2-2c^2 + \dfrac{7}{ab} + \dfrac{1}{c}&\implies\boxed{1}\\ f(b,c,a) = 2b^2+2c^2-2a^2 + \dfrac{7}{bc} + \dfrac{1}{a}&\implies\boxed{2}\\ f(c,a,b) = 2c^2+2a^2-2b^2+\dfrac{7}{ca} + \dfrac{1}{b} &\implies\boxed{3}\end{cases}$

---

$\boxed{1}=\boxed{2}:$

$2a^2+2b^2+2c^2 + \dfrac{7}{ab} + \dfrac{1}{c} = 2b^2+2c^2-2a^2 + \dfrac{7}{bc} + \dfrac{1}{a}\\ 4a^2-4c^2 = \dfrac{7}{bc} - \dfrac{7}{ab} + \dfrac{1}{a} -\dfrac{1}{c}\\ 4(a-c)(a+c) = \dfrac{7(a-c)}{abc} -\dfrac{a-c}{ac}$

Since we know that $a \neq b \neq c$ , we know that $a-c \neq 0$

Divide the equation by $(a-c)$ :

$4(a+c) = \dfrac{7}{abc} -\dfrac{1}{ac}\\ 4(a+c) = \dfrac{7-b}{abc}\\ 4abc(a+c) = 7-b\implies\boxed{4}$

---

Do the same thing for $\boxed{1} = \boxed{3}$ (Note that $b-c \neq 0$ as well):

$2a^2+2b^2+2c^2 + \dfrac{7}{ab} + \dfrac{1}{c} = 2c^2+2a^2-2b^2 + \dfrac{7}{ac} + \dfrac{1}{b}\\ 4b^2-4c^2 = \dfrac{7}{ac} - \dfrac{7}{ab} + \dfrac{1}{b} -\dfrac{1}{c}\\ 4(b-c)(b+c) = \dfrac{7(b-c)}{abc} -\dfrac{b-c}{bc}\\ 4(b+c) = \dfrac{7}{abc} -\dfrac{1}{bc}\\ 4(b+c) = \dfrac{7-a}{abc}\\ 4abc(b+c) = 7-a\implies\boxed{5}$

---

$\boxed{4} \div \boxed{5}:$

$\dfrac{4abc(a+c)}{4abc(b+c)} = \dfrac{7-b}{7-a}$

Note that for the original expression to be defined, $a,b,c \neq 0$ , because $a,b$ and $c$ are in denominators. This implies that $abc \neq 0$ as well, and we can cross off the $abc$ in the equation:

$\dfrac{a+c}{b+c} = \dfrac{7-b}{7-a}\\ (7-a)(a+c) = (7-b)(b+c)\\ 7a+7c-a^2-ac = 7b + 7c- b^2 - bc\\ 7a-7b = a^2-b^2 + ac - bc\\ (a-b)(a+b) + c(a-b) = 7(a-b)$

Remember, $a-b \neq 0$ as well, so we divide it out

$a+b+c = \color{#D61F06}{\boxed{\color{#333333}{7}}}$