Correct Answer: B To find the minimum value of the expression $P$, we will first expand and simplify it, then apply appropriate inequalities. Step 1: Expand the given expression $P$. $P = \left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)$ Expanding this product, we get: $P = 1 \cdot 1 \cdot 1 + 1 \cdot 1 \cdot \frac{1}{c} + 1 \cdot \frac{1}{b} \cdot 1 + \frac{1}{a} \cdot 1 \cdot 1 + 1 \cdot \frac{1}{b} \cdot \frac{1}{c} + \frac{1}{a} \cdot 1 \cdot \frac{1}{c} + \frac{1}{a} \cdot \frac{1}{b} \cdot 1 + \frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c}$ $P = 1 + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + \left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) + \frac{1}{abc}$ Step 2: Simplify the term $\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right)$. To combine these fractions, we find a common denominator, which is $abc$: $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{c}{abc} + \frac{a}{abc} + \frac{b}{abc} = \frac{a+b+c}{abc}$ Step 3: Utilize the given condition $a+b+c=1$. Substitute $a+b+c=1$ into the simplified term from Step 2: $\frac{a+b+c}{abc} = \frac{1}{abc}$ Step 4: Rewrite the expression $P$ with the simplified term. Now, substitute this back into the expanded expression for $P$: $P = 1 + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + \frac{1}{abc} + \frac{1}{abc}$ $P = 1 + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + \frac{2}{abc}$ Step 5: Apply the AM-HM Inequality to find the minimum of $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$. For positive real numbers $a, b, c$, the Arithmetic Mean (AM) is greater than or equal to the Harmonic Mean (HM): $\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$ Rearranging this inequality, we get: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{a+b+c}$ Given $a+b+c=1$, we substitute this value: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{1} = 9$. This minimum value of 9 is achieved when $a=b=c$. Step 6: Apply the AM-GM Inequality to find the minimum of $\frac{1}{abc}$. For positive real numbers $a, b, c$, the Arithmetic Mean (AM) is greater than or equal to the Geometric Mean (GM): $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$ Substitute $a+b+c=1$: $\frac{1}{3} \ge \sqrt[3]{abc}$ Cube both sides of the inequality: $\left(\frac{1}{3}\right)^3 \ge abc$ $\frac{1}{27} \ge abc$ Since $a, b, c$ are positive, $abc$ is positive. Taking the reciprocal of both sides reverses the inequality sign: $\frac{1}{abc} \ge 27$. This minimum value of 27 is achieved when $a=b=c$. Step 7: Substitute these minimum values back into the expression for $P$. From Step 4, we have $P = 1 + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + \frac{2}{abc}$. To find the minimum value of $P$, we use the minimum values we found for the terms: $P_{min} = 1 + \left(\text{min value of } \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) + 2 \cdot \left(\text{min value of } \frac{1}{abc}\right)$ $P_{min} = 1 + 9 + 2 \cdot 27$ $P_{min} = 1 + 9 + 54$ $P_{min} = 64$ Step 8: Verify the condition for equality. Both inequalities (AM-HM and AM-GM) achieve equality 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 $P=64$: $P = \left(1+\frac{1}{\frac{1}{3}}\right)\left(1+\frac{1}{\frac{1}{3}}\right)\left(1+\frac{1}{\frac{1}{3}}\right)$ $P = (1+3)(1+3)(1+3)$ $P = 4 \cdot 4 \cdot 4 = 64$. The minimum value of the expression is indeed 64. The final answer is $\boxed{\text{64}}$.