Inequality

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

And $(a+b+c)(ab+bc+ca) = ab^2 + a^2b + b^2c + bc^2 + ac^2 + a^2c + 3abc$

So $\dfrac{(a+b)(b+c)(c+a)}{(a+b+c)(ab+bc+ca)} = \dfrac{ab^2 + a^2b + b^2c + bc^2 + ac^2 + a^2c + 2abc}{ab^2 + a^2b + b^2c + bc^2 + ac^2 + a^2c + 3abc }$

$= 1 - \dfrac{abc}{ab^2 + a^2b + b^2c + bc^2 + ac^2 + a^2c + 3abc} = 1 -\dfrac{abc}{(a+b+c)(ab+bc+ca)}$

So, To find $N$ , we have to find maximum of $\dfrac{abc}{(a+b+c)(ab+bc+ca)}$

WLOG, $a \geq b \geq c \implies bc \leq ca \leq ab$ .

Therefore by Chebyshev's Inequality,

$(a+b+c)(bc+ca+ab) \geq 3(abc+abc+abc)$

$(a+b+c)(bc+ca+ab) \geq 9(abc)$

$\dfrac{1}{9} \geq \dfrac{abc}{(a+b+c)(bc+ca+ab)}$ .

Therefore, $N = 1 - \frac{1}{9} =\boxed{ \frac{8}{9}}$

Let:

$3u=a+b+c$

$3v^2 = ab+bc+ac$

$w^3 =abc$

By using the $u\leq v\leq w$ method,

$\frac{(a+b)(b+c)(c+a)}{(a+b+c)(ab+bc+ca)}$

$=\frac{(a+b+c)(ab+bc+ca)-abc}{(a+b+c)(ab+bc+ca)}$

$=1-\frac{w^3}{9uv^2}$

$\geq 1-\frac{w^3}{9u^3}$

$= 1-\frac{abc}{9(\frac{a+b+c}{3})^3}$

$\geq 1-\frac{abc}{9abc}$ (AM-GM)

$= \boxed{\frac{8}{9}}$

The equality holds when $a=b=c$ .

By AM-GM , we'll get (a+b)(b+c)(c+a) >= 8abc , (a+b+c) >= 3(abc)^(1/3) and (ab+bc+ca) >= 3(abc)^(2/3) so , 8abc/[(3(abc)^(1/3))(3(abc)^(2/3))] = 8/9 >= N so , a+b=8+9=17