3. Given $\log _{a} b=-2$, then the minimum value of $a+b$ is
A. $\frac{3}{2} \sqrt[3]{2}$
B. $\frac{2 \sqrt[3]{3}}{3}$
C. $\frac{3}{2} \sqrt{3}$
D. $\frac{2}{3} \sqrt{2}$
3. $\mathrm{A}$
$b=a^{-2}, a+b=a+\frac{1}{a^{2}}=\frac{a}{2}+\frac{a}{2}+\frac{1}{a^{2}} \geqslant 3 \sqrt[3]{\frac{1}{4}}=\frac{3 \sqrt[3]{2}}{2}$. And the equality can be achieved.