The image contains a mathematical inequality involving absolute values. Here’s a breakdown of the steps:

1. Given:

   • $(a, b) \in \mathbb{R}$ and $c > 0$
   • $|a| < c$ and $|b| < c$

2. Goal:

   • To prove that $\left| \frac{a + b}{2} \right| < c$

Proof Strategy:

• Since $|a| < c$, we know that $-c < a < c$.
• Similarly, $|b| < c$ implies $-c < b < c$.

Now, adding these inequalities gives:

$-c + (-c) < a + b < c + c$ $-2c < a + b < 2c$

Dividing the entire inequality by 2:

$-c < \frac{a + b}{2} < c$

Thus, $\left| \frac{a + b}{2} \right| < c$, which proves the required result.

Would you like further details on this proof or have any questions?

Here are 5 related questions to expand your understanding:

1. How would this proof change if $c \leq 0$?
2. What happens if we consider more variables, say $|a| < c_1$ and $|b| < c_2$?
3. How does this inequality relate to the concept of the midpoint between two numbers?
4. Can we generalize this result for more than two variables?
5. How would you handle the case when $a$ or $b$ are complex numbers?

Tip: When working with absolute value inequalities, breaking them down into their equivalent compound inequalities can simplify the proof process.