If x, y, z are positive real numbers such that `x+y+z=a,`then

Options

A. `(1)/(x)+(1)/(y)+(1)/(z)gt(a)/(9)`
B. `(1)/(x)+(1)/(y)+(1)/(z)gt(9)/(a)`
C. `(1)/(x)+(1)/(y)+(1)/(z)gt(9)/(a^(2))`
D. `(1)/(x)+(1)/(y)+(1)/(z)gt(a^(2))/(9)`

Explanation

To solve the problem, we need to analyze the inequality involving the positive real numbers \( x, y, z \) such that \( x + y + z = a \).

### Step-by-step Solution:

1. **Understanding the Arithmetic Mean**:
We know that the arithmetic mean of the numbers \( x, y, z \) is given by:
\[ \text{AM} = \frac{x + y + z}{3} = \frac{a}{3} \]

2. **Applying the Inequality**:
According to the inequality of arithmetic and harmonic means (AM-HM inequality), we have:
\[ \text{AM} \geq \text{HM} \]
where the harmonic mean (HM) of \( x, y, z \) is given by:
\[ \text{HM} = \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \]

3. **Setting Up the Inequality**:
From the AM-HM inequality, we can write:
\[ \frac{a}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \]

4. **Cross Multiplying**:
Cross multiplying gives us:
\[ a \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 9 \]

5. **Rearranging the Terms**:
Rearranging the above inequality leads to:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{a} \]

6. **Finding the Final Result**:
Now, we can express the inequality in terms of \( x, y, z \):
\[ 1 + \frac{1}{x} + 1 + \frac{1}{y} + 1 + \frac{1}{z} \geq 1 + \frac{9}{a} \]
Simplifying this gives:
\[ 3 + \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 3 + \frac{9}{a} \]

7. **Final Conclusion**:
Thus, we conclude that:
\[ 1 + \frac{1}{x} + 1 + \frac{1}{y} + 1 + \frac{1}{z} \geq \frac{9}{a} \]
Therefore, the correct option is:
\[ 1 + x + 1 + y + 1 + z > \frac{9}{a} \]