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The text in the image refers to the use of the **triangle inequality** and some specific properties to establish a chain of inequalities.

### Translation of the problem:

> Use the triangle inequality and the fact that \( 0 < |a| < |b| \) and \( \frac{1}{|b|} < \frac{1}{|a|} \) to establish the following chain of inequalities:
>
> \[
\left| \frac{1}{x^2 + 3} - \frac{1}{|x| + 2} \right| \leq \frac{1}{x^2 + 3} + \frac{1}{|x| + 2} \leq \frac{1}{3} + \frac{1}{2}
\]

### Understanding the problem:

- The inequality uses the **triangle inequality**, which states that for any real numbers \(a\) and \(b\), the following holds:
  \[
  |a + b| \leq |a| + |b|
  \]
- The inequality also mentions that \( 0 < |a| < |b| \) and from that, the reciprocal inequality \( \frac{1}{|b|} < \frac{1}{|a|} \) is derived. 

#### Step-by-Step Interpretation:
1. **First part:**
   - The expression starts with the absolute difference between two fractions:
     \[
     \left| \frac{1}{x^2 + 3} - \frac{1}{|x| + 2} \right|
     \]
     By the triangle inequality, we know that this absolute value of the difference is always less than or equal to the sum of the absolute values:
     \[
     \left| \frac{1}{x^2 + 3} - \frac{1}{|x| + 2} \right| \leq \frac{1}{x^2 + 3} + \frac{1}{|x| + 2}
     \]

2. **Second part:**
   - The next step is to compare this sum:
     \[
     \frac{1}{x^2 + 3} + \frac{1}{|x| + 2}
     \]
     with \( \frac{1}{3} + \frac{1}{2} \). Notice that \( \frac{1}{x^2 + 3} \leq \frac{1}{3} \) because \( x^2 + 3 \geq 3 \) for any real \(x\), and similarly \( \frac{1}{|x| + 2} \leq \frac{1}{2} \) because \( |x| + 2 \geq 2 \).

Thus, the chain of inequalities is established as:
\[
\left| \frac{1}{x^2 + 3} - \frac{1}{|x| + 2} \right| \leq \frac{1}{x^2 + 3} + \frac{1}{|x| + 2} \leq \frac{1}{3} + \frac{1}{2}
\]

### Would you like a more detailed explanation of the triangle inequality or any specific step?

Here are five related questions to consider:
1. How does the triangle inequality apply to different types of functions?
2. What is the significance of \( \frac{1}{|b|} < \frac{1}{|a|} \) in establishing inequalities?
3. How do we prove that \( \frac{1}{x^2 + 3} \leq \frac{1}{3} \)?
4. Why does \( |a| \) being less than \( |b| \) affect the inequality for their reciprocals?
5. Can the triangle inequality be generalized to higher dimensions?

**Tip:** When working with inequalities involving absolute values, always check if the triangle inequality can simplify your expression.

Utilice la desigualdad del triángulo y el hecho de que 0 < |a| < |b| entonces 1/|b| < 1/|a| para establecer la siguiente cadena de desigualdades: |(1/(x^2 + 3)) - (1/|x| + 2)| ≤ 1/(x^2 + 3) + 1/(|x| + 2) ≤ 1/3 + 1/2

Learn how to solve an inequality involving the triangle inequality and reciprocal functions. This problem focuses on proving |(1/(x^2 + 3)) - (1/|x| + 2)| ≤ 1/(x^2 + 3) + 1/(|x| + 2) ≤ 1/3 + 1/2 using algebraic concepts. Suitable for high school students.

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