Will Ravi help here?

We can write $x = q+r$ , $y = p+r$ , $z = p+q$ for $p,q,r > 0$ . Then $p+q+r=1$ , and so $x=1-p$ , $y=1-q$ , $z=1-r$ , and we are trying to extremise $(1-p)(1-q) + (1-p)(1-r) + (1-q)(1-r) - (1-p)(1-q)(1-r) \; = \; 2 - (p+q+r) + pqr \; = \; 1 + pqr$ By letting $p \to 0^+$ we see that $m=1$ . Since $pqr \; \le \; \Big(\frac{p+q+r}{3}\Big)^3 \; \le \; \frac{1}{27}$ with equality when $p=q=r=\tfrac13$ , we deduce that $n = \tfrac{28}{27}$ . Thus $m+n = \frac{55}{27},$ so $a+b = 55+27=\boxed{82}$ .

My solution using triangles properties. Let $s=(x+y+z)/2 = 1$ the triangle semiperimeter. According Lemma 8 from this paper , we have $xy + yz + zx = s^2 + (4R+r)r,$ in which $r$ and $R$ are incircle and circumcircle radius of the triangle. We also know that the triangle surface can be given by $A = sr = r = xyz/4R$ which leads us to $xyz = 4rR$ . Manipulating the expression: $E = xy + yz + zx - xyz = s^2 + (4R+r)r - 4Rr = 1 + 4Rr + r^2 - 4Rr = 1+r^2 = 1+A^2.$ The final result $E = 1 + A^2$ is equivalent to the expression $1 + pqr$ found by @Mark Hennings in the solution above. Now, we have to maximize and minimize the triangle surface. The minimum area is zero and it can be shown that a triangle with a given perimeter has the maximum possible area if it is equilateral. Then $A_{min} = 0 \to E_{min} = m = 1,$ and $A_{max} = l^2\frac{\sqrt{3}}{4} = \frac{2^2}{3^2}\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{9} \to E_{max} = n = 1 + \frac{1}{27} = \frac{28}{27}.$ Thus $m+n = 55/27$ , so $a + b = 55+27 = \boxed{82}.$