Consecutive variables

If we define $F(x,y,z) \; = \; \frac{x^2 + y^2 + z^2}{xy +xz + yz} \hspace{2cm} x \in [1,2]\,,\,y \in[2,3]\,,\, z \in[3,4]$ then $\frac{\partial F}{\partial x} \; = \; \frac{(y+z)x^2 + 2xyz - (y+z)(y^2+z^2)}{(xy + xz + yz)^2} \; = \; \frac{(y+z)(x^2 - y^2) + z(2xy - (y+z)z)}{(xy + xz + yz)^2}$ It is clear from this that $\frac{\partial F}{\partial x} \le 0$ for $x \le y \le z$ . Thus $F$ will be maximized when $x=1$ , and minimized then $x=2$ . By symmetry, we also deduce that $F$ will be maximized when $z=4$ , and minimized when $z=3$ .

To maximize $F$ , we need to maximize $g(y) \; = \; F(1,y,4) \; = \; \frac{y^2 + 17}{5y+4} \hspace{2cm} 2 \le y \le 3$ Since $g'(y) <0$ for all $2 \le y \le 3$ , we deduce that the maximum value of $F$ is $M \;= \; g(2) \; = \; F(1,2,4) \; = \; \tfrac32$ Similarly, to minimize $F$ , we need to minimize $h(y) \; = \; F(2,y,3) \; = \; \frac{y^2 + 13}{5y + 6} \hspace{2cm} 2 \le y \le 3$ Since $h'(y) \; = \; \frac{(5y-13)(y+5)}{(5y+6)^2}$ we see that $g$ is minimized when $y = \tfrac{13}{5}$ , and hence the minimum value of $F$ is $m \; = \; g(\tfrac{13}{5}) \; = \; F(2,\tfrac{13}{5},3) \; = \; \tfrac{26}{25}$ Thus $M+m = \tfrac{127}{50}$ , making the answer $\boxed{177}$ .