JBMO2020 wrote: Prove that in every triangle there are two sides with lengths $x$ and $y$ such that $(sqrt{5}-1) /2<=x/y<=(sqrt{5}+1) /2$. Prove that in every triangle there are two sides with lengths $x$ and $y$ such that $$\frac{\sqrt{5}-1}{2}\leq\frac{x}{y}\leq\frac{\sqrt{5}+1}{2}$$

Assume there are no such two sides that satisfy the given condition, $\frac{\sqrt{5}-1}{2}\leq\frac{x}{y}\leq\frac{\sqrt{5}+1}{2}$. Let the length of three sides of a triangle $a \leq b \leq c$. Since $c > \frac{\sqrt{5}+1}{2}b = b + \frac{\sqrt{5}-1}{2}b > b + a$, $a, b, c$ cannot form a triangle. It leads to contradiction, hence the problem is proved.