Call a simple graph $G$ [i]quasicolorable[/i] if we can color each edge blue, red, green, or white such that
[list]
[*] for each vertex v of degree 3 in G, the three edges incident to v are either (1) red,
green, and blue, or (2) all white,
[*] not all edges are white.
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A simple connected graph $G$ has $a$ vertices of degree $4$, $b$ vertices of degree $3$, and no other vertices, where $a$ and $b$ are positive integers. Find the smallest real number $c$ so that the following statement is true: “If $a/b > c$, then $G$ must be quasicolorable.”
To solve this problem, we need to show that if \( \frac{a}{b} > \frac{1}{4} \), then the graph \( G \) must be quasicolorable. We will proceed by considering the structure of the graph and using properties of trees and connected components.

1. **Initial Assumptions and Definitions**:
   - Let \( G \) be a simple connected graph with \( a \) vertices of degree 4 and \( b \) vertices of degree 3.
   - We need to show that if \( \frac{a}{b} > \frac{1}{4} \), then \( G \) is quasicolorable.

2. **Degree Sum Formula**:
   - The sum of the degrees of all vertices in \( G \) is given by:
     \[
     \sum \text{deg}(v) = 4a + 3b
     \]
   - Since \( G \) is a simple connected graph, the sum of the degrees is also equal to twice the number of edges \( E \):
     \[
     2|E| = 4a + 3b
     \]

3. **Edge Count and Tree Component**:
   - The number of edges in \( G \) is:
     \[
     |E| = \frac{4a + 3b}{2}
     \]
   - If \( \frac{a}{b} > \frac{1}{4} \), then \( 4a > b \).

4. **Handling Vertices of Degree 4**:
   - If there are two adjacent vertices of degree 4, we can color the edge between them with any non-white color and color all other edges white. This satisfies the quasicolorable condition since not all edges are white.

5. **Graph Induced by Degree 3 Vertices**:
   - Consider the subgraph \( G' \) induced by the \( b \) vertices of degree 3. The number of edges in \( G' \) is:
     \[
     |E(G')| \leq \frac{3b}{2}
     \]
   - Since \( 4a > b \), we have:
     \[
     \frac{4a + 3b}{2} - 4a < b
     \]
   - This implies that \( G' \) has fewer than \( b \) edges, so \( G' \) must have a connected component that is a tree \( T \).

6. **Coloring the Tree Component**:
   - Fix a vertex \( v \) in the tree \( T \) and partition the vertices of \( T \) into sets \( S_i \) where \( S_i \) consists of vertices at distance \( i \) from \( v \).
   - The sets \( S_i \) are independent sets, and edges only exist between \( S_i \) and \( S_{i+1} \).

7. **Greedy Coloring Algorithm**:
   - Use a greedy algorithm to color the edges between \( v \) and \( S_1 \) with red, green, and blue.
   - Continue this process for edges between \( S_i \) and \( S_{i+1} \), ensuring that each vertex of degree 3 has incident edges colored red, green, and blue, or all white.

8. **Coloring Edges to Degree 4 Vertices**:
   - Finally, color the edges between \( T \) and the vertices of degree 4 with non-white colors. This ensures that not all edges are white.

By following these steps, we ensure that \( G \) is quasicolorable if \( \frac{a}{b} > \frac{1}{4} \).

\(\blacksquare\)

The final answer is \( \boxed{ c = \frac{1}{4} } \)