If a, b, c, and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1 + d)?

• A) 1
• B) 4
• C) 16
• D) 64

To find the minimum value of the expression \( (1 + a)(1 + b)(1 + c)(1 + d) \) given that \( abcd = 1 \), we can use the AM-GM inequality, which states that the arithmetic mean is greater than or equal to the geometric mean.

Applying the AM-GM Inequality

We can rewrite the expression as follows:

• Let \( x_1 = 1 + a \)
• Let \( x_2 = 1 + b \)
• Let \( x_3 = 1 + c \)
• Let \( x_4 = 1 + d \)

Then, we want to minimize \( x_1 x_2 x_3 x_4 \). By the AM-GM inequality:

We have:

\( \frac{x_1 + x_2 + x_3 + x_4}{4} \geq \sqrt[4]{x_1 x_2 x_3 x_4} \)

Substituting \( x_i \) gives:

\( \frac{(1 + a) + (1 + b) + (1 + c) + (1 + d)}{4} \geq \sqrt[4]{(1 + a)(1 + b)(1 + c)(1 + d)} \)

Finding the Minimum Value

Now, since \( abcd = 1 \), we can set \( a = b = c = d = 1 \) to satisfy this condition. Thus:

• When \( a = b = c = d = 1 \), we have:
• \( (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 \times 2 \times 2 \times 2 = 16 \)

Conclusion

Therefore, the minimum value of \( (1 + a)(1 + b)(1 + c)(1 + d) \) is \( 16 \). The correct answer is:

C) 16