2014 AMC 12B Problem 13

Since $b$ is the largest of $1, a, b,$ no such triangle exists exactly when $b \ge a+1.$ Since $1$ is the largest of $\tfrac1b, \tfrac1a, 1,$ no such triangle exists exactly when $1 \ge \tfrac1a + \tfrac1b,$ that is $a \le \tfrac{b}{b-1}.$ Both conditions hold with $b$ smallest when $a+1 = b$ and $a = \tfrac{b}{b-1}$ meet, giving $b - 1 = \tfrac{b}{b-1},$ or $b^2 - 3b + 1 = 0.$ The root larger than $1$ is $b = \dfrac{3+\sqrt5}{2}.$ Thus, the correct answer is C.