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

Options

A. `(p-a)(p-b)(P-c)le(8)/(27)p^(3)`
B. `(p-a)(p-b)(P-c)ge8abc`
C. `(bc)/(a)+(ca)/(b)+(ab)/(c)gep`
D. none of these

Answer

To solve the problem, we need to analyze the given condition \( a + b + c = p \) where \( a, b, c \) are positive real numbers. We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to derive the required result.

### Step-by-Step Solution:

1. **Understanding the Condition**:
We know that \( a + b + c = p \). This means the sum of the three positive numbers \( a, b, c \) is equal to \( p \).

2. **Using AM-GM Inequality**:
According to the AM-GM inequality, for any non-negative real numbers \( x_1, x_2, \ldots, x_n \):
\[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n} \]
In our case, we can apply this to \( a, b, c \):
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
Substituting \( a + b + c = p \):
\[ \frac{p}{3} \geq \sqrt[3]{abc} \]

3. **Cubing Both Sides**:
To eliminate the cube root, we cube both sides of the inequality:
\[ \left(\frac{p}{3}\right)^3 \geq abc \]
This simplifies to:
\[ \frac{p^3}{27} \geq abc \]

4. **Rearranging the Inequality**:
Rearranging gives us:
\[ abc \leq \frac{p^3}{27} \]

5. **Conclusion**:
The inequality \( abc \leq \frac{p^3}{27} \) indicates that the product of \( a, b, c \) is maximized when \( a = b = c \). Therefore, we can conclude that under the condition \( a + b + c = p \), the maximum value of the product \( abc \) occurs when \( a, b, c \) are equal.

### Final Result:
Thus, the correct statement is:
\[ abc \leq \frac{p^3}{27} \]