A triangle $ABC$ has sides $BC = a, CA = b, AB = c$ with $a < b < c$ and area $S$. Determine the largest number $u$ and the least number $v$ such that, for every point $P$ inside $\triangle ABC$, the inequality $u \le PD + PE + PF \le v$ holds, where $D,E, F$ are the intersection points of $AP,BP,CP$ with the opposite sides.
1. **Establishing the lower bound:**
   - We start by noting that for any point \( P \) inside \( \triangle ABC \), the sum of the distances from \( P \) to the sides of the triangle, weighted by the lengths of the sides, is at least twice the area of the triangle. This can be expressed as:
     \[
     PD \cdot a + PE \cdot b + PF \cdot c \geq 2S
     \]
   - Let \( x = PD + PE + PF \). We can rewrite the inequality as:
     \[
     x \cdot c \geq 2S
     \]
   - Solving for \( x \), we get:
     \[
     x \geq \frac{2S}{c}
     \]
   - Equality holds when \( P \) is at the vertex \( C \) and \( PF \) becomes the altitude from \( C \) to \( AB \).

2. **Establishing the upper bound:**
   - The diameter of \( \triangle ABC \) is \( c \), which is the longest side of the triangle. This implies that the distance between any two points in the triangle is at most \( c \).
   - By area chasing, we have:
     \[
     \frac{PD}{DA} + \frac{PE}{EB} + \frac{PF}{FC} = 1
     \]
   - This implies that:
     \[
     \frac{x}{c} \leq 1
     \]
   - Therefore, we get:
     \[
     x \leq c
     \]
   - Equality holds when \( P \) is on the side \( AB \).

3. **Combining the results:**
   - From the above steps, we have established that:
     \[
     \frac{2S}{c} \leq PD + PE + PF \leq c
     \]

The final answer is \( \boxed{\frac{2S}{c} \leq PD + PE + PF \leq c} \)