已知 $a$ 和 $b$ 都是正数常数，在同一幅图上，草拟以下函数的图像：
$y = \frac{2}{|x - a|} \quad \text{and} \quad y = b|x - a|.$ 由此或通过其他方法，解不等式 $\frac{2}{|x - a|} > b|x - a|$。

1. Determine the shape and key features of $y = \frac{2}{|x - a|}$.

The expression $|x - a|$ is always positive for $x \neq a$, so $y = \frac{2}{|x - a|}$ is strictly positive.
As $x \to a$, the denominator approaches $0$, so $y \to \infty$, meaning $x = a$ is a vertical asymptote.
As $x \to \pm\infty$, $y \to 0$, meaning the $x$-axis ($y = 0$) is a horizontal asymptote.
At $x = 0$, the $y$-intercept is $y = \frac{2}{|-a|} = \frac{2}{a}$ (since $a > 0$).

2. Determine the shape and key features of $y = b|x - a|$.

The function $y = b|x - a|$ is a modulus graph, which forms a 'V' shape opening upwards (since $b > 0$).
The minimum point (vertex) occurs when $|x - a| = 0$, giving the coordinate $(a, 0)$.
At $x = 0$, the $y$-intercept is $y = b|-a| = ab$.

3. Sketch both graphs on the same coordinate axes.

Draw the $x$ and $y$ axes.
Mark $x = a$ on the positive $x$-axis and draw a vertical dotted line to represent the asymptote and axis of symmetry.
Plot the vertex $(a, 0)$ and draw the straight V-shape lines for $y = b|x - a|$, labelling the $y$-intercept $ab$.
Draw the two symmetric branches of $y = \frac{2}{|x - a|}$ approaching $x = a$ upwards and the $x$-axis outwards, labelling the $y$-intercept $\frac{2}{a}$.
Ensure the graphs intersect exactly twice, once on each side of $x = a$.

1. Multiply both sides of the inequality by $|x - a|$ to eliminate the denominator, noting the condition for validity.

For the expression $\frac{2}{|x - a|}$ to be defined, $x \neq a$.
For $x \neq a$, we know that $|x - a| > 0$.
Since $|x - a|$ is strictly positive, we can multiply both sides of the inequality by $|x - a|$ without reversing the inequality sign:
$\frac{2}{|x - a|} > b|x - a|$
$2 > b|x - a|^2$

2. Solve the resulting inequality for $x - a$.

Since $b$ is a positive constant ($b > 0$), we divide by $b$:
$|x - a|^2 < \frac{2}{b}$
Since $|x - a|^2 = (x - a)^2$ and the square root of both sides gives $|x - a|$:
$|x - a| < \sqrt{\frac{2}{b}}$
Using the property of modulus inequalities ($|X| < k \iff -k < X < k$):
$-\sqrt{\frac{2}{b}} < x - a < \sqrt{\frac{2}{b}}$
$a - \sqrt{\frac{2}{b}} < x < a + \sqrt{\frac{2}{b}}$

3. State the final solution set, excluding the value where the original expression is undefined.

The original inequality is undefined at $x = a$.
Therefore, we must explicitly exclude $x = a$ from the solution interval.
$a - \sqrt{\frac{2}{b}} < x < a \quad \text{or} \quad a < x < a + \sqrt{\frac{2}{b}}$