Inequalities

Table of Contents

• Reciprocals and Inequalities
  • Removing the Absolute Value
  • When to flip inequality signs
    • Examples
    • Exception
• Proof of the Reverse Triangle Inequality

Reciprocals and Inequalities

$$\frac{d}{dx} [ln(x)] = \frac{1}{x}$$

Removing the Absolute Value

• If $|x| < a$, then $-a < x < a$.
• If $|x| > a$, then $x < -a$ or $x > a$.

When to flip inequality signs

• Main Rule: When taking the reciprocal of an inequality (e.g., $a < b$), you must flip the inequality sign ($\frac{1}{a} > \frac{1}{b}$).

• Key Condition: This rule only works if both sides have the same sign.

Examples

• Both Positive:

  • $3 < 10 \implies \frac{1}{3} > \frac{1}{10}$

• Both Negative:

  • $-10 < -3 \implies -\frac{1}{10} > -\frac{1}{3}$

Exception

• If the signs are different, the inequality sign does not flip.
  • $-2 < 5 \implies -\frac{1}{2} < \frac{1}{5}$ (A negative is always less than a positive).

Proof of the Reverse Triangle Inequality

The proof relies on the standard triangle inequality, which states that for any real numbers $a$ and $b$, $|a + b| \le |a| + |b|$.

1. First Step

   We start by writing $x$ in a convenient form: $x = (x - y) + y$.

   Now, we take the absolute value of both sides and apply the triangle inequality:

   $|x| = |(x - y) + y| \le |x - y| + |y|$

   By subtracting $|y|$ from both sides, we get our first result:

   $|x| - |y| \le |x - y| \qquad \text{(1)}$

2. Second Step

   Similarly, we can start by writing $y$ as $y = (y - x) + x$.

   Again, we take the absolute value and apply the triangle inequality:

   $|y| = |(y - x) + x| \le |y - x| + |x|$

   Subtracting $|x|$ from both sides gives:

   $|y| - |x| \le |y - x|$

   Since $|y - x| = |-(x - y)| = |x - y|$, we can rewrite this as:

   $|y| - |x| \le |x - y|$

   Multiplying the entire inequality by $-1$ reverses the inequality sign:

   $-(|y| - |x|) \ge -|x - y|$

   $|x| - |y| \ge -|x - y| \qquad \text{(2)}$

3. Conclusion

   We can now combine inequalities (1) and (2) into a single statement:

   $-|x - y| \le |x| - |y| \le |x - y|$

   This compound inequality is the definition of the absolute value. Therefore, we can conclude: $| |x| - |y| | \le |x - y|$

   This completes the proof. ✅