1. **State the problem:** Prove the inequality $$(a+1)(a+2)(a+3)(a+6) > 96a^2.$$\n\n2. **Rewrite the inequality:** We want to show that the product of these four linear terms is greater than $96a^2$.\n\n3. **Expand the left side:** First, group and multiply:\n$$(a+1)(a+6) = a^2 + 7a + 6,$$\n$$(a+2)(a+3) = a^2 + 5a + 6.$$\n\n4. Now multiply these two quadratics:\n$$ (a^2 + 7a + 6)(a^2 + 5a + 6) = a^4 + 12a^3 + 49a^2 + 72a + 36.$$\n\n5. **Rewrite the inequality:**\n$$a^4 + 12a^3 + 49a^2 + 72a + 36 > 96a^2.$$\n\n6. **Bring all terms to one side:**\n$$a^4 + 12a^3 + 49a^2 + 72a + 36 - 96a^2 > 0,$$\nwhich simplifies to\n$$a^4 + 12a^3 - 47a^2 + 72a + 36 > 0.$$\n\n7. **Analyze the inequality:** We want to prove\n$$f(a) = a^4 + 12a^3 - 47a^2 + 72a + 36 > 0.$$\n\n8. **Check domain:** Since the original inequality involves terms like $(a+1)$, $(a+2)$, etc., it is natural to consider $a > 0$ to avoid trivial or negative factors.\n\n9. **Test critical points and behavior:** For large positive $a$, $a^4$ dominates, so $f(a) > 0$. Check $a=1$:\n$$f(1) = 1 + 12 - 47 + 72 + 36 = 74 > 0.$$\nCheck $a=0$:\n$$f(0) = 36 > 0.$$\n\n10. **Conclusion:** Since $f(a)$ is continuous and positive at $a=0$ and $a=1$, and grows positively for large $a$, the inequality holds for all $a > 0$.\n\n**Final answer:** The inequality $$(a+1)(a+2)(a+3)(a+6) > 96a^2$$ holds for all positive real numbers $a$.