Triangle Inequality

Note that $0 \le |u+v| \le |u|+|v|$ for all reals $u,v$ (the first inequality is by definition of norm, the second inequality is the triangle inequality). Thus $0 \le \frac{|u+v|}{|u|+|v|} \le 1$ for all nonzero reals $u,v$ , so each term of the sum is in the range $[0,1]$ . This means $E \le 1+1+1 = 3$ ; indeed, since we can obtain $E = 3$ by, for example, $x = y = z = 1$ , we obtain $b = 3$ .

To obtain $a$ , we need to proceed further. By pigeonhole principle, at least two of the three of $x,y,z$ have the same sign; without loss of generality, $x,y$ are positive. (If the two with the same sign are not $x,y$ ; rename the variables; if they are negative, take the additive inverse of all elements, which will retain the value of $E$ .) Since $x,y$ are positive, so as $x+y$ ; thus, $|x| = x, |y| = y, |x+y| = x+y$ , and so the first term of the sum is $\frac{x+y}{x+y} = 1$ . Thus $E \ge 1+0+0 = 1$ , and indeed we can achieve a minimum of 1 by, for example, $x = y = 1, z = -1$ . Thus $a = 1$ , and so $a+b = 1+3 = \boxed{4}$ .