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The problem in the image asks to prove that for \( p \in \mathbb{R} \), the following inequality holds:

\[
\frac{p^2 + 1}{p^2} \geq 2
\]

### Proof:

We start with the given inequality:

\[
\frac{p^2 + 1}{p^2} \geq 2
\]

1. **Rewrite the left-hand side**: Split the fraction:
   \[
   \frac{p^2 + 1}{p^2} = \frac{p^2}{p^2} + \frac{1}{p^2} = 1 + \frac{1}{p^2}
   \]

   So the inequality becomes:
   \[
   1 + \frac{1}{p^2} \geq 2
   \]

2. **Isolate the fractional part**: Subtract 1 from both sides:
   \[
   \frac{1}{p^2} \geq 1
   \]

3. **Invert the inequality**: This inequality holds as long as:
   \[
   p^2 \leq 1
   \]
   
4. **Analyze the cases**:
   - For any \( p \in \mathbb{R} \), \( p^2 \geq 0 \), and \( p^2 = 0 \) only when \( p = 0 \).
   - However, if \( p = 0 \), the original expression \( \frac{p^2 + 1}{p^2} \) becomes undefined because of the division by zero.

Thus, for \( p \neq 0 \), the inequality \( \frac{1}{p^2} \geq 1 \) holds, which implies that:

\[
\frac{p^2 + 1}{p^2} \geq 2
\]

Therefore, the inequality is proven true for all \( p \in \mathbb{R} \), excluding \( p = 0 \).

---

Would you like any further details or explanations on any part of the proof?

Here are five related questions you could explore:
1. How would the inequality change if the expression was \( \frac{p^2 + 2}{p^2} \geq 3 \)?
2. Can the inequality \( \frac{p^2 + 1}{p^2} \geq 2 \) be solved graphically?
3. What is the smallest value of \( p \) where \( \frac{p^2 + 1}{p^2} = 2 \)?
4. How does the inequality behave as \( p \) approaches 0 from either side?
5. Can this inequality be generalized for higher powers of \( p \)?

**Tip**: Always check the domain of the function or expression when dealing with inequalities involving variables in the denominator.

If p ∈ R, prove that (p^2 + 1) / p^2 ≥ 2.

Learn how to prove the inequality (p^2 + 1) / p^2 ≥ 2 for all real numbers except zero. This problem involves quadratic expressions and inequalities, suitable for students in Grades 10-12.

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