3. For arbitrary real numbers $a$ and $b(b \neq 0)$, find the minimum value of the expression $a^{2}+b^{2}+\frac{a}{b}+\frac{1}{b^{2}}$.
Answer: $\sqrt{3}$

Solution. Since

$$
a^{2}+b^{2}+\frac{a}{b}+\frac{1}{b^{2}}=\left(a+\frac{1}{2 b}\right)^{2}+b^{2}+\frac{3}{4 b^{2}} \geqslant b^{2}+\frac{3}{4 b^{2}} \geqslant 2 \sqrt{b^{2} \cdot \frac{3}{4 b^{2}}}=\sqrt{3}
$$

the expression $a^{2}+b^{2}+\frac{a}{b}+\frac{1}{b^{2}}$ is always no less than $\sqrt{3}$. Equality is achieved if $a+\frac{1}{2 b}=0$ and $b^{2}=\frac{3}{4 b^{2}}$, i.e., when $b=\frac{\sqrt[4]{3}}{\sqrt{2}}$ and $a=-\frac{1}{\sqrt[4]{3} \sqrt{2}}$.