If a, b, c are positive real numbers such that `a+b+c=2` then, which one of the following is true?

Options

A. `(2-a)(2-b)(2-c)ge8abc`
B. `(1)/(a)+(1)/(b)+(1)/(c)ge2`
C. `(2-a)(2-b)(2-c)lt8abc`
D. `(1)/(a)+(1)/(b)+(1)/( c )=2`

Answer

To solve the problem, we need to analyze the given condition \( a + b + c = 2 \) and apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We know that \( a, b, c \) are positive real numbers such that \( a + b + c = 2 \).

2. **Applying AM-GM Inequality**:
We will apply the AM-GM inequality to pairs of variables.

- For \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
Since \( a + b = 2 - c \), we can rewrite this as:
\[ \frac{2 - c}{2} \geq \sqrt{ab} \]
Multiplying both sides by 2 gives:
\[ 2 - c \geq 2\sqrt{ab} \quad \text{(1)} \]

- For \( b \) and \( c \):
\[ \frac{b + c}{2} \geq \sqrt{bc} \]
Here, \( b + c = 2 - a \), so:
\[ \frac{2 - a}{2} \geq \sqrt{bc} \]
Multiplying both sides by 2 gives:
\[ 2 - a \geq 2\sqrt{bc} \quad \text{(2)} \]

- For \( a \) and \( c \):
\[ \frac{a + c}{2} \geq \sqrt{ac} \]
Here, \( a + c = 2 - b \), so:
\[ \frac{2 - b}{2} \geq \sqrt{ac} \]
Multiplying both sides by 2 gives:
\[ 2 - b \geq 2\sqrt{ac} \quad \text{(3)} \]

3. **Combining the Results**:
Now we have three inequalities:
\[ 2 - c \geq 2\sqrt{ab} \quad (1) \]
\[ 2 - a \geq 2\sqrt{bc} \quad (2) \]
\[ 2 - b \geq 2\sqrt{ac} \quad (3) \]

4. **Multiplying the Inequalities**:
We will multiply all three inequalities:
\[ (2 - c)(2 - a)(2 - b) \geq 8\sqrt{abc} \]

5. **Final Relation**:
The final relation we obtain is:
\[ (2 - a)(2 - b)(2 - c) \geq 8abc \]

6. **Conclusion**:
This inequality gives us a relationship between the sums and products of \( a, b, c \). Therefore, we can conclude that the correct option is the one that states this relationship.