Prove the inequality $$\frac{a + b}{a^2 -ab + b^2} \le \frac{4}{ |a + b|}$$ for all real numbers $a$ and $b$ with $a + b\ne 0$. When does equality hold?
To prove the inequality 
\[ \frac{a + b}{a^2 - ab + b^2} \le \frac{4}{|a + b|} \]
for all real numbers \(a\) and \(b\) with \(a + b \ne 0\), we will start by analyzing the given expression and transforming it to a more manageable form.

1. **Rewrite the inequality:**
   \[ \frac{a + b}{a^2 - ab + b^2} \le \frac{4}{|a + b|} \]

2. **Multiply both sides by \(|a + b|\):**
   \[ \frac{(a + b)^2}{a^2 - ab + b^2} \le 4 \]

3. **Let \(s = a + b\) and \(p = ab\):**
   The inequality becomes:
   \[ \frac{s^2}{a^2 - ab + b^2} \le 4 \]

4. **Express \(a^2 - ab + b^2\) in terms of \(s\) and \(p\):**
   \[ a^2 - ab + b^2 = (a + b)^2 - 3ab = s^2 - 3p \]

5. **Substitute \(s\) and \(p\) into the inequality:**
   \[ \frac{s^2}{s^2 - 3p} \le 4 \]

6. **Rearrange the inequality:**
   \[ s^2 \le 4(s^2 - 3p) \]
   \[ s^2 \le 4s^2 - 12p \]
   \[ 3s^2 \le 12p \]
   \[ p \ge \frac{s^2}{4} \]

7. **Analyze the condition for equality:**
   Equality holds if and only if:
   \[ \frac{a + b}{a^2 - ab + b^2} = \frac{4}{|a + b|} \]
   This implies:
   \[ \frac{s^2}{s^2 - 3p} = 4 \]
   \[ s^2 = 4(s^2 - 3p) \]
   \[ s^2 = 4s^2 - 12p \]
   \[ 3s^2 = 12p \]
   \[ p = \frac{s^2}{4} \]

   Since \(p = ab\) and \(s = a + b\), we have:
   \[ ab = \frac{(a + b)^2}{4} \]

   This is a quadratic equation in \(a\) and \(b\):
   \[ ab = \frac{a^2 + 2ab + b^2}{4} \]
   \[ 4ab = a^2 + 2ab + b^2 \]
   \[ 2ab = a^2 + b^2 \]
   \[ a^2 - 2ab + b^2 = 0 \]
   \[ (a - b)^2 = 0 \]
   \[ a = b \]

   Therefore, equality holds when \(a = b\).

The final answer is \( \boxed{ a = b } \).