If a, b, c are positive integers, then
`((a^(2)+b^(2)+c^(2))/(a+b+c))^(a+b+c)gta^(x)b^(y)c^(z)`, then

Options

A. `x=a, y=b, z=c`
B. `x=b, y=a, z=c`
C. `x=(1)/(a),y=(1)/(b),z=(1)/( c )`
D. `x=y=z=1`

Explanation

To solve the inequality \(\left(\frac{a^2 + b^2 + c^2}{a + b + c}\right)^{(a + b + c)} > a^x b^y c^z\) for positive integers \(a\), \(b\), and \(c\), we will use the concept of weighted arithmetic mean and weighted geometric mean.

### Step-by-step Solution:

1. **Understanding Weighted Means**:
- The weighted arithmetic mean (WAM) of \(a\), \(b\), and \(c\) with weights \(a\), \(b\), and \(c\) respectively is given by:
\[ \text{WAM} = \frac{a \cdot a + b \cdot b + c \cdot c}{a + b + c} = \frac{a^2 + b^2 + c^2}{a + b + c} \]
- The weighted geometric mean (WGM) is given by:
\[ \text{WGM} = \left(a^a b^b c^c\right)^{\frac{1}{a + b + c}} \]

2. **Applying the Inequality**:
- By the property of weighted means, we know that:
\[ \text{WAM} \geq \text{WGM} \]
- Thus, we can write:
\[ \frac{a^2 + b^2 + c^2}{a + b + c} \geq \left(a^a b^b c^c\right)^{\frac{1}{a + b + c}} \]

3. **Raising Both Sides to the Power of \(a + b + c\)**:
- Raising both sides of the inequality to the power of \(a + b + c\) gives:
\[ \left(\frac{a^2 + b^2 + c^2}{a + b + c}\right)^{(a + b + c)} \geq a^a b^b c^c \]

4. **Comparing with the Given Expression**:
- The inequality we need to satisfy is:
\[ \left(\frac{a^2 + b^2 + c^2}{a + b + c}\right)^{(a + b + c)} > a^x b^y c^z \]
- From the previous step, we have:
\[ a^a b^b c^c \geq a^x b^y c^z \]

5. **Identifying Values of \(x\), \(y\), and \(z\)**:
- To satisfy the inequality \(a^a b^b c^c \geq a^x b^y c^z\), we can compare the exponents:
- This leads to:
- \(x \leq a\)
- \(y \leq b\)
- \(z \leq c\)
- The maximum values that satisfy this are:
- \(x = a\)
- \(y = b\)
- \(z = c\)

### Conclusion:
Thus, the values of \(x\), \(y\), and \(z\) are:
\[ \boxed{x = a, \; y = b, \; z = c} \]