If a, b, c and d are four positive numbers such that `a+b+c+d=4`, then what is the maximum value of `(a+1)(b+1)(c+1)(d+1)`?

Options

A. 32
B. 8
C. 16
D. 81

Answer

To find the maximum value of \((a+1)(b+1)(c+1)(d+1)\) given that \(a + b + c + d = 4\) and \(a, b, c, d\) are positive numbers, we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step 1: Rewrite the expression
We start by rewriting the expression we want to maximize:
\[ (a+1)(b+1)(c+1)(d+1) = abcd + (abc + abd + acd + bcd) + (ab + ac + ad + bc + bd + cd) + (a + b + c + d) + 1 \]

### Step 2: Apply AM-GM Inequality
Using the AM-GM inequality, we know that for any non-negative numbers \(x_1, x_2, x_3, x_4\):
\[ \frac{x_1 + x_2 + x_3 + x_4}{4} \geq \sqrt[4]{x_1 x_2 x_3 x_4} \]
In our case, we can apply this to \(a+1, b+1, c+1, d+1\):
\[ \frac{(a+1) + (b+1) + (c+1) + (d+1)}{4} \geq \sqrt[4]{(a+1)(b+1)(c+1)(d+1)} \]

### Step 3: Simplify the left-hand side
Calculating the left-hand side:
\[ (a+1) + (b+1) + (c+1) + (d+1) = (a + b + c + d) + 4 = 4 + 4 = 8 \]
Thus, we have:
\[ \frac{8}{4} = 2 \geq \sqrt[4]{(a+1)(b+1)(c+1)(d+1)} \]

### Step 4: Raise both sides to the fourth power
Raising both sides to the fourth power gives:
\[ 2^4 \geq (a+1)(b+1)(c+1)(d+1) \]
This simplifies to:
\[ 16 \geq (a+1)(b+1)(c+1)(d+1) \]

### Step 5: Achieving equality
The equality in the AM-GM inequality holds when all the terms are equal, which means:
\[ a+1 = b+1 = c+1 = d+1 \]
This implies:
\[ a = b = c = d \]
Since \(a + b + c + d = 4\), we have:
\[ 4a = 4 \implies a = 1 \]
Thus, \(b = 1\), \(c = 1\), and \(d = 1\).

### Step 6: Calculate the maximum value
Now substituting back:
\[ (a+1)(b+1)(c+1)(d+1) = (1+1)(1+1)(1+1)(1+1) = 2 \times 2 \times 2 \times 2 = 16 \]

### Conclusion
The maximum value of \((a+1)(b+1)(c+1)(d+1)\) is \(\boxed{16}\).

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