$a/b < (1 + \sqrt{5})/2 \iff a^2 - ab - b^2 < 0$?

If $b=0, a^2<0$ which is impossible

So, $b\ne0, b^2>0$ consequently, $$a^2-ab-b^2<0\iff\left(\dfrac ab\right)^2-\left(\dfrac ab\right)-1<0$$

Now the roots of $x^2-x-1=0$ are $$x=\dfrac{1\pm\sqrt5}2$$

We can prove if $(x-a)(x-b)<0$ with $a<b;$

$$a<x<b$$