Croatian Mathematical Society Competitions

Let $x=a/b$ and $y=b/a$. Since $a,b>0$, $x,y>0$ and $xy=1$.

We are given: $$\frac{a}{b}+\frac{b}{a}=x+y=3.$$ Since $xy=1$, $x$ and $y$ are the roots of $t^2-3t+1=0$.

Now, $$\frac{a^2}{b}+\frac{b^2}{a}=\frac{a^2}{b}+\frac{b^2}{a}=a\cdot \frac{a}{b}+b\cdot \frac{b}{a}=ax+by.$$ But this is not immediately helpful. Instead, let us try expressing everything in terms of $a$ and $b$.

Let $S=a+b$, $P=ab$.

We have: $$\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=3⟹a^2+b^2=3ab.$$ Also, $$\frac{a^2}{b}+\frac{b^2}{a}=\frac{a^3+b^3}{ab}=10.$$ So $a^3+b^3=10ab$.

Recall that $a^3+b^3=(a+b{)}^3-3ab(a+b)=S^3-3PS$.

So: $$S^3-3PS=10P$$ But from above, $a^2+b^2=3ab⟹(a+b{)}^2-2ab=3ab⟹S^2-2P=3P⟹S^2=5P$.

Now, substitute $P=S^2/5$ into the cubic equation: $$S^3-3PS=10P$$ $$S^3-3PS-10P=0$$ $$S^3-3SP-10P=0$$ $$S^3-(3S+10)P=0$$ $$S^3=(3S+10)P$$ But $P=S^2/5$, so: $$S^3=(3S+10)\cdot \frac{S^2}{5}$$ $$S^3=\frac{(3S+10)S^2}{5}$$ $$5S^3=(3S+10)S^2$$ $$5S^3-3S^3-10S^2=0$$ $$2S^3-10S^2=0$$ $$2S^3=10S^2$$ $$S^3=5S^2$$ Since $S>0$, $S=5$.

Now, $P=S^2/5=25/5=5$.

Therefore, $a$ and $b$ are the roots of $t^2-St+P=0$, i.e. $$t^2-5t+5=0$$ So $$a,b=\frac{5\pm \sqrt{25-20}}{2}=\frac{5\pm \sqrt{5}}{2}$$

Now, $$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=\frac{S}{P}=\frac{5}{5}=1.$$

Answer: $$\boxed{1}$$