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 $3a^2\equiv 1(\mathrm{mod}5)$, which is impossible since $3a^2\equiv 0(\mathrm{mod}5)$, $3a^2\equiv 3(\mathrm{mod}5)$ for $a\equiv \pm 1(\mathrm{mod}5)$, and $3a^2\equiv 12\equiv 2(\mathrm{mod}5)$ for $a\equiv \pm 2(\mathrm{mod}5)$.

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

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