Find the largest positive integer $n$ for which the inequality 
\[ \frac{a+b+c}{abc+1}+\sqrt[n]{abc} \leq \frac{5}{2}\]
holds true for all $a, b, c \in [0,1]$. Here we make the convention $\sqrt[1]{abc}=abc$.
To find the largest positive integer \( n \) for which the inequality 
\[ \frac{a+b+c}{abc+1}+\sqrt[n]{abc} \leq \frac{5}{2} \]
holds true for all \( a, b, c \in [0,1] \), we will analyze the critical points and simplify the problem step by step.

1. **Identify Critical Points:**
   By using methods such as Lagrange multipliers, we can identify the critical points where the inequality might be tight. The critical points to check are:
   - \( a = b = c = \sqrt[3]{x} \) for some \( x \in [0,1] \)
   - \( c = 1 \), \( a = b = \sqrt{x} \) for some \( x \in [0,1] \)
   - \( b = c = 1 \), \( a = x \) for some \( x \in [0,1] \)

2. **Evaluate at Critical Points:**
   - For \( a = b = c = \sqrt[3]{x} \):
     \[
     \text{LHS} = \frac{3\sqrt[3]{x}}{x+1} + \sqrt[n]{x}
     \]
   - For \( c = 1 \), \( a = b = \sqrt{x} \):
     \[
     \text{LHS} = \frac{2\sqrt{x} + 1}{x+1} + \sqrt[n]{x}
     \]
   - For \( b = c = 1 \), \( a = x \):
     \[
     \text{LHS} = \frac{x + 2}{x+1} + \sqrt[n]{x}
     \]

3. **Simplify and Compare:**
   - Note that \( 2\sqrt{x} + 1 \leq x + 2 \) since it is equivalent to \( (\sqrt{x} - 1)^2 \geq 0 \).
   - Also, \( 3\sqrt[3]{x} \leq 2\sqrt{x} + 1 \) since it is equivalent to \( (\sqrt[6]{x} - 1)^2(2\sqrt[6]{x} + 1) \geq 0 \).

   Therefore, the most critical expression to check is:
   \[
   \frac{x + 2}{x + 1} + \sqrt[n]{x} \leq \frac{5}{2}
   \]
   This simplifies to:
   \[
   \frac{1}{x + 1} + \sqrt[n]{x} \leq \frac{3}{2}
   \]

4. **Check for \( n = 3 \):**
   Let \( t = \sqrt[3]{x} \). The inequality becomes:
   \[
   \frac{1}{t^3 + 1} + t \leq \frac{3}{2}
   \]
   This can be rewritten as:
   \[
   2t^4 - 3t^3 + 2t - 1 \leq 0
   \]
   Factoring, we get:
   \[
   (t - 1)(2t^3 - t^2 - t + 1) \leq 0
   \]
   Since \( 0 \leq t \leq 1 \) and \( 2t^3 - t^2 - t + 1 > 0 \) for \( t \in [0,1] \), the inequality holds true.

5. **Check for \( n = 4 \):**
   Let \( t = \sqrt[4]{x} \). The inequality becomes:
   \[
   \frac{1}{t^4 + 1} + t \leq \frac{3}{2}
   \]
   This can be rewritten as:
   \[
   2t^5 - 3t^4 + 2t - 1 \leq 0
   \]
   Factoring, we get:
   \[
   (t - 1)^2(2t^3 + t^2 - 1) \leq 0
   \]
   This is clearly false for \( t < 1 \) but \( t \) close enough to 1.

Therefore, the largest such \( n \) is \( \boxed{3} \).