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.