Prove that if a, 6, and c are positive real numbers with ab = c, then a < yc or 6 < Vc.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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can someone answer this asap, please

 

**Prove that if \(a\), \(b\), and \(c\) are positive real numbers with \(ab = c\), then \(a \leq \sqrt{c}\) or \(b \leq \sqrt{c}\).**

To prove this, let's analyze the given condition \(ab = c\).

1. Consider the contrapositive: suppose both \(a > \sqrt{c}\) and \(b > \sqrt{c}\).

2. Then we have:
   \[
   a > \sqrt{c} \quad \text{and} \quad b > \sqrt{c}
   \]

3. Multiplying these inequalities gives:
   \[
   ab > \sqrt{c} \cdot \sqrt{c} = c
   \]

4. However, this contradicts the given condition \(ab = c\).

5. Therefore, it must be the case that \(a \leq \sqrt{c}\) or \(b \leq \sqrt{c}\).

This proves the required inequality.
Transcribed Image Text:**Prove that if \(a\), \(b\), and \(c\) are positive real numbers with \(ab = c\), then \(a \leq \sqrt{c}\) or \(b \leq \sqrt{c}\).** To prove this, let's analyze the given condition \(ab = c\). 1. Consider the contrapositive: suppose both \(a > \sqrt{c}\) and \(b > \sqrt{c}\). 2. Then we have: \[ a > \sqrt{c} \quad \text{and} \quad b > \sqrt{c} \] 3. Multiplying these inequalities gives: \[ ab > \sqrt{c} \cdot \sqrt{c} = c \] 4. However, this contradicts the given condition \(ab = c\). 5. Therefore, it must be the case that \(a \leq \sqrt{c}\) or \(b \leq \sqrt{c}\). This proves the required inequality.
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