The correct answer is C. Let the given expression be $E$. We have: $E = (1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})$ First, expand the expression: $E = \left(\frac{a+1}{a}\right) \left(\frac{b+1}{b}\right) \left(\frac{c+1}{c}\right)$ $E = \frac{(a+1)(b+1)(c+1)}{abc}$ Now, expand the numerator: $(a+1)(b+1)(c+1) = abc + (ab+bc+ca) + (a+b+c) + 1$ Substitute the given condition $a+b+c=1$ into the expanded numerator: $(a+1)(b+1)(c+1) = abc + (ab+bc+ca) + 1 + 1$ $(a+1)(b+1)(c+1) = abc + (ab+bc+ca) + 2$ Substitute this back into the expression for $E$: $E = \frac{abc + (ab+bc+ca) + 2}{abc}$ $E = 1 + \frac{ab+bc+ca}{abc} + \frac{2}{abc}$ $E = 1 + \left(\frac{ab}{abc} + \frac{bc}{abc} + \frac{ca}{abc}\right) + \frac{2}{abc}$ $E = 1 + \left(\frac{1}{c} + \frac{1}{a} + \frac{1}{b}\right) + \frac{2}{abc}$ Now, we need to find the minimum value of this expression using inequalities, specifically the AM-GM inequality and its variants. **Step 1: Apply AM-GM to $a, b, c$ to find the minimum of $\frac{1}{abc}$** For positive real numbers $a, b, c$, the AM-GM inequality states: $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$ Given $a+b+c=1$, we have: $\frac{1}{3} \ge \sqrt[3]{abc}$ Cube both sides: $\left(\frac{1}{3}\right)^3 \ge abc$ $\frac{1}{27} \ge abc$ This implies: $\frac{1}{abc} \ge 27$ **Step 2: Apply a variant of AM-GM (or Cauchy-Schwarz) to find the minimum of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$** For positive real numbers $a, b, c$, we know that: $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})(a+b+c) \ge (1+1+1)^2 = 9$ Given $a+b+c=1$, we substitute this into the inequality: $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})(1) \ge 9$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge 9$ **Step 3: Substitute the minimum values back into the expression $E$** We have $E = 1 + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + \frac{2}{abc}$. Using the minimum values found in Step 1 and Step 2: $E \ge 1 + (9) + 2(27)$ $E \ge 1 + 9 + 54$ $E \ge 64$ **Step 4: Check for equality condition** The equality in AM-GM and the Cauchy-Schwarz variant holds when $a=b=c$. Since $a+b+c=1$, this implies $a=b=c=\frac{1}{3}$. Let's check if $a=b=c=\frac{1}{3}$ yields 64: $E = \left(1+\frac{1}{1/3}\right) \left(1+\frac{1}{1/3}\right) \left(1+\frac{1}{1/3}\right)$ $E = (1+3)(1+3)(1+3)$ $E = 4 \times 4 \times 4 = 64$ Since the minimum value is achieved when $a=b=c=1/3$, the minimum value of the expression is 64. The final answer is $\boxed{\text{64}}$