To solve the inequality:

\[
\frac{1}{|x+2|} < \frac{1}{|x+1|}
\]

We will approach it step by step.

### Step 1: Analyze the inequality

The inequality compares the reciprocals of two absolute value expressions. Since the reciprocal function \( \frac{1}{y} \) is a decreasing function for positive \( y \), the inequality \( \frac{1}{|x+2|} < \frac{1}{|x+1|} \) implies:

\[
|x+2| > |x+1|
\]

### Step 2: Consider the different cases for absolute values

The behavior of absolute values changes based on the sign of the expression inside. We need to consider the cases where \( x+2 \) and \( x+1 \) change sign.

1. **Case 1:** \( x \geq -1 \)
2. **Case 2:** \( -2 \leq x < -1 \)
3. **Case 3:** \( x < -2 \)

### Case 1: \( x \geq -1 \)

In this case:
\[
|x+2| = x+2 \quad \text{and} \quad |x+1| = x+1
\]

The inequality becomes:
\[
x+2 > x+1
\]

This simplifies to:
\[
2 > 1
\]

Which is always true. Thus, for \( x \geq -1 \), the inequality holds.

### Case 2: \( -2 \leq x < -1 \)

In this case:
\[
|x+2| = x+2 \quad \text{and} \quad |x+1| = -(x+1) = -x-1
\]

The inequality becomes:
\[
x+2 > -x-1
\]

Simplifying this:
\[
2x > -3 \quad \Rightarrow \quad x > -\frac{3}{2}
\]

Since \( x > -\frac{3}{2} \) must also satisfy \( -2 \leq x < -1 \), the valid interval for this case is:
\[
-\frac{3}{2} < x < -1
\]

### Case 3: \( x < -2 \)

In this case:
\[
|x+2| = -(x+2) = -x-2 \quad \text{and} \quad |x+1| = -(x+1) = -x-1
\]

The inequality becomes:
\[
-x-2 > -x-1
\]

This simplifies to:
\[
-2 > -1
\]

Which is false. Thus, for \( x < -2 \), the inequality does not hold.

### Step 3: Combine the results

From the above cases:

- For \( x \geq -1 \), the inequality holds for all \( x \).
- For \( -\frac{3}{2} < x < -1 \), the inequality also holds.
- For \( x < -2 \), the inequality does not hold.

### Final Answer

The solution to the inequality is:

\[
x > -\frac{3}{2}
\]

### Do you have any questions or need further details?

Here are some related questions:

1. What are the steps to solve absolute value inequalities?
2. How does the sign of \( x \) affect the solution of inequalities involving absolute values?
3. What is the role of case analysis in solving inequalities with absolute values?
4. How do you graph the solution of the inequality on a number line?
5. Can this approach be applied to more complex inequalities involving absolute values?

**Tip:** When dealing with absolute values in inequalities, always consider the different cases based on where the expression inside the absolute value changes sign.

Learn how to solve the inequality 1/(|x+2|) < 1/(|x+1|) step-by-step with detailed explanations. This problem involves inequalities and absolute value properties, suitable for high school students.

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