Let $M = \{(a,b,c)\in R^3 :0 <a,b,c<\frac12$  with  $a+b+c=1 \}$ and  $f: M\to  R$ given as  $$f(a,b,c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{abc}$$
Find the best (real) bounds $\alpha$ and $\beta$ such that $f(M) = \{f(a,b,c): (a,b,c)\in M\}\subseteq [\alpha,\beta]$ and determine whether any of them is achievable.
To find the best bounds $\alpha$ and $\beta$ for the function $f(a,b,c) = 4\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) - \frac{1}{abc}$ over the set $M = \{(a,b,c) \in \mathbb{R}^3 : 0 < a, b, c < \frac{1}{2}, a + b + c = 1\}$, we need to analyze the behavior of $f$ on the boundary and within the interior of $M$.

1. **Boundary Analysis:**
   - Consider the boundary of the set $\overline{M} = \{(a,b,c) \in \mathbb{R}^3 : 0 \le a, b, c \le \frac{1}{2}, a + b + c = 1\}$.
   - On the boundary, at least one of $a$, $b$, or $c$ is either $0$ or $\frac{1}{2}$.
   - Since $a, b, c > 0$ in $M$, we only need to consider the cases where $a, b, c \to 0$ or $a, b, c \to \frac{1}{2}$.

2. **Case 1: $a \to 0$, $b, c \to \frac{1}{2}$:**
   - As $a \to 0$, $\frac{1}{a} \to \infty$.
   - Therefore, $f(a,b,c) \to 4\left(\frac{1}{0} + \frac{1}{\frac{1}{2}} + \frac{1}{\frac{1}{2}}\right) - \frac{1}{0 \cdot \frac{1}{2} \cdot \frac{1}{2}} \to \infty$.
   - This case is not possible since $a, b, c > 0$.

3. **Case 2: $a \to \frac{1}{2}$, $b, c \to \frac{1}{4}$:**
   - As $a \to \frac{1}{2}$, $b, c \to \frac{1}{4}$, we have:
     \[
     f\left(\frac{1}{2}, \frac{1}{4}, \frac{1}{4}\right) = 4\left(2 + 4 + 4\right) - \frac{1}{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}} = 4 \cdot 10 - 32 = 40 - 32 = 8.
     \]
   - Therefore, $f \to 8$.

4. **Interior Analysis using Lagrange Multipliers:**
   - We use Lagrange multipliers to find the critical points of $f$ subject to the constraint $a + b + c = 1$.
   - Define $g(a, b, c) = a + b + c - 1$.
   - The Lagrangian is:
     \[
     \mathcal{L}(a, b, c, \lambda) = 4\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) - \frac{1}{abc} + \lambda (a + b + c - 1).
     \]
   - Taking partial derivatives and setting them to zero:
     \[
     \frac{\partial \mathcal{L}}{\partial a} = -\frac{4}{a^2} + \frac{1}{a^2bc} + \lambda = 0,
     \]
     \[
     \frac{\partial \mathcal{L}}{\partial b} = -\frac{4}{b^2} + \frac{1}{ab^2c} + \lambda = 0,
     \]
     \[
     \frac{\partial \mathcal{L}}{\partial c} = -\frac{4}{c^2} + \frac{1}{abc^2} + \lambda = 0,
     \]
     \[
     \frac{\partial \mathcal{L}}{\partial \lambda} = a + b + c - 1 = 0.
     \]
   - Solving these equations, we find that $a = b = c = \frac{1}{3}$.
   - Substituting $a = b = c = \frac{1}{3}$ into $f$:
     \[
     f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = 4\left(3 + 3 + 3\right) - \frac{1}{\left(\frac{1}{3}\right)^3} = 4 \cdot 9 - 27 = 36 - 27 = 9.
     \]

5. **Conclusion:**
   - The minimum value of $f$ is $8$ and the maximum value is $9$.
   - Both bounds are achievable.

The final answer is $\boxed{[8, 9]}$