Let's solve the inequality:

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

### Step 1: Understand the inequality
Since we are dealing with fractions, the inequality \( \frac{1}{a} < \frac{1}{b} \) (where \( a > 0 \) and \( b > 0 \)) is equivalent to \( a > b \). Thus, the inequality can be rewritten as:

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

### Step 2: Consider the cases based on the definition of absolute values

#### Case 1: \( x \geq -1 \)
For \( x \geq -1 \), we have:
\[
|x+2| = x+2 \quad \text{and} \quad |x+1| = x+1
\]
So the inequality becomes:
\[
x+2 > x+1
\]
Subtracting \( x \) from both sides:
\[
2 > 1
\]
This is always true, so for \( x \geq -1 \), the inequality holds.

#### Case 2: \( -2 \leq x < -1 \)
For \( -2 \leq x < -1 \), we have:
\[
|x+2| = x+2 \quad \text{and} \quad |x+1| = -(x+1)
\]
So the inequality becomes:
\[
x+2 > -(x+1)
\]
Simplifying further:
\[
x + 2 > -x - 1
\]
Adding \( x \) to both sides:
\[
2x + 2 > -1
\]
Subtracting 2 from both sides:
\[
2x > -3
\]
Dividing by 2:
\[
x > -\frac{3}{2}
\]
Since we are considering \( -2 \leq x < -1 \), the solution in this case is \( -\frac{3}{2} < x < -1 \).

#### Case 3: \( x < -2 \)
For \( x < -2 \), we have:
\[
|x+2| = -(x+2) \quad \text{and} \quad |x+1| = -(x+1)
\]
So the inequality becomes:
\[
-(x+2) > -(x+1)
\]
Simplifying:
\[
-x-2 > -x-1
\]
Adding \( x \) to both sides:
\[
-2 > -1
\]
This is not true, so there is no solution in this case.

### Final Answer
Combining the results from the three cases:

1. For \( x \geq -1 \), the inequality holds for all \( x \geq -1 \).
2. For \( -2 \leq x < -1 \), the inequality holds for \( -\frac{3}{2} < x < -1 \).
3. For \( x < -2 \), there is no solution.

Thus, the solution set is:

\[
x \in \left(-\frac{3}{2}, -1\right) \cup \left[-1, \infty\right)
\]

Would you like more details or have any questions?

Here are some related questions for further exploration:
1. How would the solution change if the inequality were reversed?
2. What if the absolute values were replaced with another type of function?
3. How does the solution change if you add constants to the absolute value expressions?
4. What is the impact of multiplying the entire inequality by \(|x+1|\) or \(|x+2|\)?
5. Can you solve a similar inequality where one side is a polynomial instead of a fraction?

**Tip:** When solving inequalities involving absolute values, breaking the problem into different cases based on the sign of the expressions inside the absolute values can simplify the process.

Learn how to solve the fractional inequality 1/(|x+2|) < 1/(|x+1|) step-by-step with detailed explanations. This problem involves inequalities, absolute values, and fractional expressions. Suitable for high school students studying algebra.

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