3. For arbitrary real numbers $a$ and $b (a \neq 0)$, find the minimum value of the expression $\frac{1}{a^{2}}+2 a^{2}+3 b^{2}+4 a b$.
Answer: $\sqrt{\frac{8}{3}}$

Since

$$
\frac{1}{a^{2}}+2 a^{2}+3 b^{2}+4 a b=\left(\sqrt{3} b+\frac{2 a}{\sqrt{3}}\right)^{2}+\frac{2 a^{2}}{3}+\frac{1}{a^{2}} \geqslant \frac{2 a^{2}}{3}+\frac{1}{a^{2}} \geqslant 2 \sqrt{\frac{2 a^{2}}{3} \cdot \frac{1}{a^{2}}}=\sqrt{\frac{8}{3}}
$$

the expression $\frac{1}{a^{2}}+2 a^{2}+3 b^{2}+4 a b$ is always no less than $\sqrt{\frac{8}{3}}$. Equality is achieved if $\sqrt{3} b+\frac{2 a}{\sqrt{3}}=0$ and $\frac{2 a^{2}}{3}=\frac{1}{a^{2}}$, i.e., when $a=\sqrt[4]{\frac{3}{2}}$ and $b=-\left(\frac{2}{3}\right)^{3 / 4}$.