Team selection tests for JBMO 2018

Let $a,b,c$ be the side lengths of the given triangle and we may assume that $a\ge b\ge c$. Denote $m=\frac{a}{b}$ and $n=\frac{b}{c}$. Note that $$m,n\ge 1>\frac{\sqrt{5}-1}{2}.$$ 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 $n+1>mn$. Hence, $$n+1>mn>\frac{\sqrt{5}+1}{2}n,\text{ or }n<\frac{\sqrt{5}+1}{2},$$ 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. $\square$