9. Let $a+b=1, b>0, a \neq 0$, then the minimum value of $\frac{1}{|a|}+\frac{2|a|}{b}$ is

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Solve $2 \sqrt{2}-1$.
Analysis $\frac{1}{|a|}+\frac{2|a|}{b}=\frac{a+b}{|a|}+\frac{2|a|}{b}=\frac{a}{|a|}+\left(\frac{b}{|a|}+\frac{2|a|}{b}\right) \geq \frac{a}{|a|}+2 \sqrt{\frac{b}{|a|} \cdot \frac{2|a|}{b}}=2 \sqrt{2}+\frac{a}{|a|}$, where the equality holds when $\frac{b}{|a|}=\frac{2|a|}{b}$, i.e., $b^{2}=2 a^{2}$.
When $a>0$, $\frac{1}{|a|}+\frac{2|a|}{b} \geq 2 \sqrt{2}+1$, the minimum value is reached when $a=\sqrt{2}-1, b=\sqrt{2} a=2-\sqrt{2}$; when $a<0$, $\frac{1}{|a|}+\frac{2|a|}{b} \geq 2 \sqrt{2}-1$, the minimum value $2 \sqrt{2}-1$ is reached when $a=-(\sqrt{2}+1), b=-\sqrt{2} a$.