1. **Problem Statement:** Given that $\tan \theta = \frac{a^2 + ab + b^2}{ab}$, show that $\tan \theta > 3$. 2. **Rewrite the expression:** Start with the given expression: $$\tan \theta = \frac{a^2 + ab + b^2}{ab}$$ 3. **Divide each term in the numerator by $ab$:** $$\tan \theta = \frac{a^2}{ab} + \frac{ab}{ab} + \frac{b^2}{ab} = \frac{a}{b} + 1 + \frac{b}{a}$$ 4. **Introduce a variable substitution:** Let $x = \frac{a}{b}$. Then, $$\tan \theta = x + 1 + \frac{1}{x}$$ 5. **Analyze the expression $x + 1 + \frac{1}{x}$ for positive $x$:** Since $a$ and $b$ are real numbers and appear in denominators, assume $x > 0$. 6. **Use the AM-GM inequality:** Arithmetic Mean-Geometric Mean inequality states: $$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1$$ Multiplying both sides by 2, $$x + \frac{1}{x} \geq 2$$ 7. **Add 1 to both sides:** $$x + 1 + \frac{1}{x} \geq 2 + 1 = 3$$ 8. **Check equality conditions:** Equality holds if and only if $x = \frac{1}{x} \Rightarrow x = 1$, i.e., $a = b$. 9. **Conclusion:** Since $a$ and $b$ are positive real numbers, $$\tan \theta = x + 1 + \frac{1}{x} \geq 3$$ Therefore, $$\tan \theta > 3$$ except when $a = b$ where it equals 3. **Final answer:** $\boxed{\tan \theta \geq 3}$ with strict inequality unless $a = b$.