After swimming around the ocean with some snorkling gear, Joshua walks back to the beach where Alexis works on a mural in the sand beside where they drew out symbol lists.  Joshua walks directly over the mural without paying any attention.

"You're a square, Josh."

"No, $\textit{you're}$ a square," retorts Joshua.  "In fact, you're a $\textit{cube}$, which is $50\%$ freakier than a square by dimension.  And before you tell me I'm a hypercube, I'll remind you that mom and dad confirmed that they could not have given birth to a four dimension being."

"Okay, you're a cubist caricature of male immaturity," asserts Alexis.

Knowing nothing about cubism, Joshua decides to ignore Alexis and walk to where he stashed his belongings by a beach umbrella.  He starts thinking about cubes and computes some sums of cubes, and some cubes of sums: \begin{align*}1^3+1^3+1^3&=3,\\1^3+1^3+2^3&=10,\\1^3+2^3+2^3&=17,\\2^3+2^3+2^3&=24,\\1^3+1^3+3^3&=29,\\1^3+2^3+3^3&=36,\\(1+1+1)^3&=27,\\(1+1+2)^3&=64,\\(1+2+2)^3&=125,\\(2+2+2)^3&=216,\\(1+1+3)^3&=125,\\(1+2+3)^3&=216.\end{align*}  Josh recognizes that the cubes of the sums are always larger than the sum of cubes of positive integers.  For instance, 

\begin{align*}(1+2+4)^3&=1^3+2^3+4^3+3(1^2\cdot 2+1^2\cdot 4+2^2\cdot 1+2^2\cdot 4+4^2\cdot 1+4^2\cdot 2)+6(1\cdot 2\cdot 4)\\&>1^3+2^3+4^3.\end{align*}

Josh begins to wonder if there is a smallest value of $n$ such that \[(a+b+c)^3\leq n(a^3+b^3+c^3)\] for all natural numbers $a$, $b$, and $c$.  Joshua thinks he has an answer, but doesn't know how to prove it, so he takes it to Michael who confirms Joshua's answer with a proof.  What is the correct value of $n$ that Joshua found?
To find the smallest value of \( n \) such that \((a+b+c)^3 \leq n(a^3 + b^3 + c^3)\) for all natural numbers \( a \), \( b \), and \( c \), we start by expanding the left-hand side and comparing it to the right-hand side.

1. **Expand \((a+b+c)^3\):**
   \[
   (a+b+c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc
   \]

2. **Compare the expanded form to \( n(a^3 + b^3 + c^3) \):**
   We need to find \( n \) such that:
   \[
   a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \leq n(a^3 + b^3 + c^3)
   \]

3. **Simplify the inequality:**
   \[
   3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \leq (n-1)(a^3 + b^3 + c^3)
   \]

4. **Test specific values to find the smallest \( n \):**
   - Let \( a = b = c = 1 \):
     \[
     (1+1+1)^3 = 27 \quad \text{and} \quad 1^3 + 1^3 + 1^3 = 3
     \]
     \[
     27 \leq n \cdot 3 \implies n \geq 9
     \]

   - Let \( a = 1, b = 1, c = 0 \):
     \[
     (1+1+0)^3 = 8 \quad \text{and} \quad 1^3 + 1^3 + 0^3 = 2
     \]
     \[
     8 \leq n \cdot 2 \implies n \geq 4
     \]

   - Let \( a = 1, b = 0, c = 0 \):
     \[
     (1+0+0)^3 = 1 \quad \text{and} \quad 1^3 + 0^3 + 0^3 = 1
     \]
     \[
     1 \leq n \cdot 1 \implies n \geq 1
     \]

5. **Conclusion:**
   The smallest \( n \) that satisfies all these conditions is \( n = 9 \). This is because the most restrictive condition comes from the case where \( a = b = c = 1 \).

The final answer is \( \boxed{9} \).