Such Big Denominator you got there!

$abc=1$ so $1\ge a, 1\ge b, 1\ge c$

Implies that the maximum value of $a+b+c$ is $3$ Where $a=b=c=1$

Note that:

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

= $\frac { 2 }{ { (3) }^{ 2 } }$ = $\frac { 2 }{ 9 }$

So $x=2$ and $y=9$

$x+y=11$

$\dfrac { 2 }{ 2ab+2ac+2bc+a^{ 2 }+b^{ 2 }+c^{ 2 } }$

For is maxima we need to calculate minima of $2ab+2ac+2bc+a^{ 2 }+b^{ 2 }+c^{ 2 }$

By AM-GM

$\Rightarrow \dfrac { ab+bc+ca }{ 3 } \geq \sqrt [ 3 ]{ (abc)^{ 2 } } \\ \Rightarrow ab+bc+ca\geq 1\times 3\\ \Rightarrow 2ab+2bc+2ca\geq 6$

Again by AM GM

$\Rightarrow \dfrac { a^{ 2 }+b^{ 2 }+c^{ 2 } }{ 3 } \geq \sqrt [ 3 ]{ (abc)^{ 2 } } \\ \Rightarrow \dfrac { a^{ 2 }+b^{ 2 }+c^{ 2 } }{ 3 } \geq 1\\ \quad \Rightarrow a^{ 2 }+b^{ 2 }+c^{ 2 }\geq 3\\ \Rightarrow \dfrac { 2 }{ 2ab+2ac+2bc+a^{ 2 }+b^{ 2 }+c^{ 2 } } \leq \dfrac { 2 }{ 6+3 } \\ \\ \Rightarrow \dfrac { 2 }{ 2ab+2ac+2bc+a^{ 2 }+b^{ 2 }+c^{ 2 } } \leq \dfrac { 2 }{ 9 }$

$\color{royalblue}{\huge\Rightarrow\boxed{x+y=11}}$

First, note that the expression $\dfrac{a+b+c}{abc} = \dfrac{1}{ab} + \dfrac{1}{ac} + \dfrac{1}{bc}$ . Then, multiply by $a+b+c$ again to obtain $\dfrac{a+b+c}{ab} + \dfrac{a+b+c}{ac} + \dfrac{a+b+c}{bc} = \dfrac{1}{b} + \dfrac{1}{a} + \dfrac{c}{ab} + \dfrac{1}{c} + \dfrac{b}{ac} + \dfrac{1}{a} + \dfrac{a}{bc} + \dfrac{1}{c} + \dfrac{1}{b} = \dfrac{2}{a} + \dfrac{2}{b} + \dfrac{2}{c} + \dfrac{a}{bc} + \dfrac{b}{ac} + \dfrac{c}{ab}$ . We take the reciprocal of this exression to obtain $\dfrac{1}{\dfrac{2}{a} + \dfrac{2}{b} + \dfrac{2}{c} + \dfrac{a}{bc} + \dfrac{b}{ac} + \dfrac{c}{ab}}$ , and mulitply by $\dfrac{1}{abc} = \dfrac{1}{1}$ to obtain $\dfrac{1}{2bc+2ab+2ac+a^2+b^2+c^2}$ , then multiply by $2$ to obtain the desired expression. Thus, our expression is equal to $\dfrac{2}{(a+b+c)^2}$ , where $a+b+c=3$ , Thus, we have $\dfrac{2}{9}$ , and $2+9=\boxed{11}$ .

We know that A.M. > G.M.

So (a+b+c)/3 > 1 which gives a+b+c is always greater than or equal to 3. Thus least value of a+b+c is 3

Now 2 / (2ab +2bc + 2ca + a^2 + b^2 + c^2) = 2/(a+b+c)^2 thus its max value is possible when a+b+c have its least value i.e. 3

So max value of expression is 2/9

therefore 2+9 = 11

(2ab + 2ac +2bc + a^{2} +b^{2} +c^{2}) =( \boxed{(a + b +c )^{2}) and min value of( \boxed{(a + b +c )^{2}) is 3