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)`.

Options

A. 1
B. 4
C. 16
D. 64

Explanation

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 (Arithmetic Mean-Geometric Mean) inequality.

### Step-by-Step Solution:

1. **Rewrite the expression**:
We start with the expression:
\[ (1+a)(1+b)(1+c)(1+d) \]

2. **Expand the expression**:
Expanding this expression gives us:
\[ = 1 + (a+b+c+d) + (ab + ac + ad + bc + bd + cd) + (abc + abd + acd + bcd) + abcd \]
Since \( abcd = 1 \), we can substitute this into the expression:
\[ = 1 + (a+b+c+d) + (ab + ac + ad + bc + bd + cd) + (abc + abd + acd + bcd) + 1 \]
Thus, we have:
\[ = 2 + (a+b+c+d) + (ab + ac + ad + bc + bd + cd) + (abc + abd + acd + bcd) \]

3. **Apply AM-GM inequality**:
By the AM-GM inequality:
\[ \frac{a+b+c+d}{4} \geq \sqrt[4]{abcd} = \sqrt[4]{1} = 1 \]
Therefore,
\[ a+b+c+d \geq 4 \]

4. **Consider pairs of variables**:
Similarly, for the products of pairs:
\[ ab + ac + ad + bc + bd + cd \geq 6 \cdot \sqrt[6]{(abcd)^{3}} = 6 \cdot \sqrt[6]{1} = 6 \]
And for the products of triples:
\[ abc + abd + acd + bcd \geq 4 \cdot \sqrt[4]{(abcd)^{3}} = 4 \cdot \sqrt[4]{1} = 4 \]

5. **Combine the inequalities**:
Now we can combine all these inequalities:
\[ (a+b+c+d) + (ab + ac + ad + bc + bd + cd) + (abc + abd + acd + bcd) \geq 4 + 6 + 4 = 14 \]

6. **Final calculation**:
Adding this to the constant term:
\[ (1+a)(1+b)(1+c)(1+d) \geq 2 + 14 = 16 \]

7. **Conclusion**:
Therefore, the minimum value of \( (1+a)(1+b)(1+c)(1+d) \) is:
\[ \boxed{16} \]