, 2018

Let $a, b, c$ be the side lengths of the given triangle and we may assume that $a \geq b \geq c$. Denote $m=\frac{a}{b}$ and $n=\frac{b}{c}$. Note that
$$m, n \geq 1 > \frac{\sqrt{5}-1}{2} .$$
We will show that at least one of $m, n$ is less than or equal to $\frac{\sqrt{5}+1}{2}$. Assume that both $m, n > \frac{\sqrt{5}+1}{2}$. Since $b + c > a$, it follows that $n + 1 > m n$. Hence,
$$n + 1 > m n > \frac{\sqrt{5}+1}{2} n, \text{ or } n < \frac{\sqrt{5}+1}{2},$$
which is a contradiction. Therefore, one of two numbers $m, n$ is less than or equal $\frac{\sqrt{5}+1}{2}$, this finishes our proof. $\square$