Any triangle must satisfy

This problem was originally published in Crux Mathematicorum in 1986.

If we define $S \; = \; S(x,y,z) \; = \; \frac{y-z}{y+z} + \frac{x-y}{x+y} + \frac{z-x}{z+x} \; = \; \frac{(y-z)(x-y)(x-z)}{(y+z)(x+y)(x+z)}$ then $S$ is a antisymmetric function of $x,y,z$ , so that (to determine the supremum of $|S|$ ), it is sufficient to restrict attention to the case $x \ge y \ge z$ , when $S \ge 0$ . If we write $p \; = \; y - z \ge 0 \qquad q \; = \; x - y \ge 0 \qquad r \; = \; y + z - x > 0$ then $S \; = \; F(p,q,r) \; = \; \frac{pq(p+q)}{(p+2q+2r)(2p+3q+2r)(p+3q+2r)}$ and we are interested in finding the supremum of $F(p,q,r)$ for $p,q \ge 0$ and $r > 0$ . Since $F(0,q,r) = F(p,0,r)= 0$ , we might as well find the supremum of $F(p,q,r)$ for $p,q,r > 0$ . Since $F(p,q,r)$ is homogeneous in $p,q,r$ , we may assume without loss of generality that $q=1$ , and it is clear that $F(p,1,r) \; \le \; F(p,1,0) \; = \; \lim_{r \to 1-}F(p,1,r) \; = \; \frac{p(p+1)}{(p+2)(2p+3)(p+3)}$ Looking for turning points, we have $\frac{d}{dp}F(p,1,0) \; = \; -2 \frac{p^4 + 2p^3 - 7p^2 - 18p - 9}{(p + 2)^2 (p + 3)^2 (2p + 3)^2}$ Thus we need to solve $\begin{aligned} p^4 + 2p^3 - 7p^2 - 18p - 9 & = 0 \\ (p^2 + p + 1)^2 & = 10(p + 1)^2 \\ \big[p^2 + (1 + \sqrt{10})p + 1 + \sqrt{10}\big]\big[p^2 + (1 - \sqrt{10})p + 1 - \sqrt{10}\big] & = 0 \\ p^2 + (1 + \sqrt{10})p + 1 + \sqrt{10} \; = \; 0 \qquad \text{or} \qquad & p^2 + (1 - \sqrt{10})p + 1 - \sqrt{10}\big] \;=\; 0 \end{aligned}$ The first of these quadratics has no positive roots, and so, solving the second, we obtain $\begin{aligned} \big(p + \tfrac12(1 - \sqrt{10})\big)^2 & = \; \tfrac14(7 + 2\sqrt{10}) \; = \; \tfrac14(\sqrt{2} + \sqrt{5})^2 \\ p & = \tfrac12(1 \pm \sqrt{2})(1 \pm \sqrt{5}) - 1 \end{aligned}$ Since we want the positive root, we deduce that $p \; = \; \tfrac12(1 + \sqrt{2})(1 + \sqrt{5}) - 1$ Thus we deduce that $S \; \le \; F(p,1,0) \; \le \; F\big(\tfrac12(1 + \sqrt{2})(1 + \sqrt{5}) - 1,1,0) \; = \; \frac{8\sqrt{2} - 5\sqrt{5}}{3}$ making the answer $8+2+5+5+3 = \boxed{23}$ .