Show that for any positive real numbers $a,b,c$ the following inequality is true:
$$4(a^3+b^3+c^3+3)\geq 3(a+1)(b+1)(c+1)$$
When does equality hold?
To prove the inequality for any positive real numbers \(a, b, c\):
\[ 4(a^3 + b^3 + c^3 + 3) \geq 3(a+1)(b+1)(c+1) \]

1. **Expand the right-hand side:**
   \[
   3(a+1)(b+1)(c+1) = 3(abc + ab + ac + bc + a + b + c + 1)
   \]

2. **Rewrite the inequality:**
   \[
   4(a^3 + b^3 + c^3 + 3) \geq 3(abc + ab + ac + bc + a + b + c + 1)
   \]

3. **Simplify the inequality:**
   \[
   4a^3 + 4b^3 + 4c^3 + 12 \geq 3abc + 3ab + 3ac + 3bc + 3a + 3b + 3c + 3
   \]

4. **Rearrange the terms:**
   \[
   4a^3 + 4b^3 + 4c^3 + 9 \geq 3abc + 3ab + 3ac + 3bc + 3a + 3b + 3c
   \]

5. **Use the AM-GM inequality:**
   By the AM-GM inequality, we know that for any non-negative real numbers \(x_1, x_2, \ldots, x_n\),
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   Applying this to \(a^3, b^3, c^3\), we get:
   \[
   a^3 + b^3 + c^3 \geq 3\sqrt[3]{a^3 b^3 c^3} = 3abc
   \]
   Therefore,
   \[
   4(a^3 + b^3 + c^3) \geq 4 \cdot 3abc = 12abc
   \]

6. **Combine the inequalities:**
   \[
   4a^3 + 4b^3 + 4c^3 + 9 \geq 12abc + 9
   \]
   Since \(12abc \geq 3abc + 3ab + 3ac + 3bc + 3a + 3b + 3c\), we have:
   \[
   4a^3 + 4b^3 + 4c^3 + 9 \geq 3abc + 3ab + 3ac + 3bc + 3a + 3b + 3c
   \]

7. **Equality condition:**
   Equality holds when \(a = b = c = 1\). In this case,
   \[
   4(1^3 + 1^3 + 1^3 + 3) = 4(6) = 24
   \]
   and
   \[
   3(1+1)(1+1)(1+1) = 3 \cdot 2 \cdot 2 \cdot 2 = 24
   \]
   Thus, equality holds when \(a = b = c = 1\).

The final answer is \( \boxed{ 4(a^3 + b^3 + c^3 + 3) \geq 3(a+1)(b+1)(c+1) } \) and equality holds when \(a = b = c = 1\).