If `a+b+c=6` then the maximum value of `sqrt(4a+1)+sqrt(4b+1)+sqrt(4c+1)=`

Options

A. 9
B. 6
C. 4
D. 12

Explanation

To find the maximum value of \( \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} \) given the constraint \( a + b + c = 6 \), we can follow these steps:

### Step 1: Express the terms in a simpler form
We start with the expression:
\[ \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} \]
We can rewrite each term:
\[ \sqrt{4a+1} = \sqrt{4a} + \sqrt{1} = 2\sqrt{a} + 1 \]
Thus, we can express the entire sum as:
\[ \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} = (2\sqrt{a} + 1) + (2\sqrt{b} + 1) + (2\sqrt{c} + 1) \]
This simplifies to:
\[ 2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + 3 \]

### Step 2: Use the Cauchy-Schwarz inequality
To maximize \( \sqrt{a} + \sqrt{b} + \sqrt{c} \), we apply the Cauchy-Schwarz inequality:
\[ (\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \leq (1 + 1 + 1)(a + b + c) \]
Substituting \( a + b + c = 6 \):
\[ (\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \leq 3 \times 6 = 18 \]
Taking the square root of both sides gives:
\[ \sqrt{a} + \sqrt{b} + \sqrt{c} \leq \sqrt{18} = 3\sqrt{2} \]

### Step 3: Substitute back to find the maximum value
Now substituting back into our expression:
\[ 2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + 3 \leq 2(3\sqrt{2}) + 3 = 6\sqrt{2} + 3 \]

### Step 4: Find the maximum achievable value
To achieve equality in the Cauchy-Schwarz inequality, \( a, b, c \) must be equal. Therefore, let:
\[ a = b = c = 2 \]
Then:
\[ \sqrt{4a + 1} = \sqrt{4(2) + 1} = \sqrt{9} = 3 \]
Thus:
\[ \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} = 3 + 3 + 3 = 9 \]

### Conclusion
The maximum value of \( \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} \) given \( a + b + c = 6 \) is:
\[ \boxed{9} \]