Soar high!

Let $f(x,y,z) = x^2y + y^2z + z^2x$ . Note that $f(x,y,z) - f(x,z,y) \; = \; (x-y)(x-z)(y-z)$ Since we can permute $x,y,z$ cyclically without changing $f(x,y,z)$ , we can assume without loss of generality that $x \ge y,z \ge 0$ . The above identity then shows us that the maximum value of $f(x,y,z)$ will be obtained when $x \ge y \ge z \ge 0$ (since, in that case, $f(x,y,z) - f(x,z,y) \ge 0$ ). Since $f(x+z,y,0) - f(x,y,z) \; = \; xz(y-z) + (x-y)(y-z) + yz^2 \; \ge\; 0 \hspace{2cm} x \ge y \ge z \ge 0\;,$ we see that the largest value of $f(x,y,z)$ is obtained when $x \ge y \ge z = 0$ . Thus we simply need to maximize $f(x,1-x,0) \; = \; x^2(1-x) \hspace{2cm} 0 \le x \le 1$ and this maximum is achieved when $x=\tfrac23$ , yielding a maximum of $\tfrac{4}{27}$ . This makes the answer $\boxed{31}$ .