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,
$$\frac{a^2}{b} + \frac{b^2}{a} = \frac{a^2}{b} + \frac{b^2}{a} = a \cdot \frac{a}{b} + b \cdot \frac{b}{a} = a x + b y.$$
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:
$$\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab} = 3 \implies a^2 + b^2 = 3ab.$$
Also,
$$\frac{a^2}{b} + \frac{b^2}{a} = \frac{a^3 + b^3}{ab} = 10.$$
So $a^3 + b^3 = 10ab$.

Recall that $a^3 + b^3 = (a + b)^3 - 3ab(a + b) = S^3 - 3P S$.

So:
$$S^3 - 3P S = 10P$$
But from above, $a^2 + b^2 = 3ab \implies (a + b)^2 - 2ab = 3ab \implies S^2 - 2P = 3P \implies S^2 = 5P$.

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

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

Therefore, $a$ and $b$ are the roots of $t^2 - S t + P = 0$, i.e.
$$t^2 - 5 t + 5 = 0$$
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:
$$\boxed{1}$$