Belarusian Mathematical Olympiad

Multiplying the left-hand side of the required inequality by $a\sqrt{3} + b\sqrt{5}$, we obtain $|a\sqrt{3} - b\sqrt{5}|(a\sqrt{3} + b\sqrt{5}) = |3a^2 - 5b^2|$. If $|3a^2 - 5b^2| = 1$ for some integers $a$ and $b$, then either $3a^2 - 5b^2 = 1$ or $3a^2 - 5b^2 = -1$. In the first case we have $3a^2 \equiv 1 \pmod{5}$, which is impossible since $3a^2 \equiv 0 \pmod{5}$, $3a^2 \equiv 3 \pmod{5}$ for $a \equiv \pm 1 \pmod{5}$, and $3a^2 \equiv 12 \equiv 2 \pmod{5}$ for $a \equiv \pm 2 \pmod{5}$.

In the second case we have $-5b^2 \equiv (-1) \pmod{3}$, i.e. $5b^2 \equiv 1 \pmod{3}$, which is impossible since $5b^2 \equiv 0 \pmod{3}$ for $b \equiv 0 \pmod{3}$, and $5b^2 \equiv 5 \pmod{3}$ for $b \equiv \pm 1$.

Therefore, $|3a^2 - 5b^2| \neq 1$ for all integers $a$ and $b$, hence $|3a^2 - 5b^2| \geq 2$ for all integers $a$ and $b$. So
$$\begin{aligned} |a\sqrt{3} - b\sqrt{5}| &= \frac{|3a^2 - 5b^2|}{a\sqrt{3} + b\sqrt{5}} \ge \frac{2}{a\sqrt{3} + b\sqrt{5}} = \frac{4}{2a\sqrt{3} + 2b\sqrt{5}} \\ &\ge [2\sqrt{3} < 4,\ 2\sqrt{5} < 5] \ge \frac{4}{4a + 5b} \end{aligned}$$
as required.