Triple Inequalities

Given $x,y,z > 0$ with $xy + xz + yz = 5$ :

If (WILOG) $x \le y \le z$ then $x^2 \le y^2 \le z^2$ and $z^{-3} \le y^{-3} \le x^{-3}$ , and hence $\frac{x^2}{y^3} + \frac{y^2}{z^3} + \frac{z^2}{y^3} \; \ge \; x^{-1} + y^{-1} + z^{-1} \; = \; \frac{5}{xyz}$ By the AM/GM inequality, $(xyz)^{\frac23} \; \le \; \frac{xy + xz + yz}{3} = \tfrac53$ and hence, putting these together, $\frac{x^2}{y^3} + \frac{y^2}{z^3} + \frac{z^2}{x^3} \; \ge \; 3\sqrt{\tfrac35}$ with equality when $x=y=z$ . Since $(x-y)^2 + (x-z)^2 + (y-z)^2 \ge 0$ we deduce that $x^2 + y^2 + z^2 \ge xy + xz + yz$ and equality occurs when $x=y=z$ . Moreover, $(x+y+z)^2 \ge 3(xy + xz + yz) = 15$ so that $x + y + z \ge \sqrt{15}$ , with equality when $x=y=z$ . Thus $\frac{x^2}{y^3} + \frac{y^2}{z^3} + \frac{z^2}{x^3} + x + y + z \; \ge \; 3\sqrt{\tfrac35} + \sqrt{15} \; = \; \frac{ 8 \sqrt{15} } { 5}$ with equality when $x=y=z$ . Thus $a=8$ , $b=15$ , $c=5$ , making the answer $8+15+5= \boxed{28}$ .