If `x > 0, y > 0, z>0 and x + y + z = 1` then the minimum value of `x/(2-x)+y/(2-y)+z/(2-z)` is:

Options

A. (a) `0.2`
B. (b) `0.4`
C. (c) `0.6`
D. (d) `0.8`

Explanation

To find the minimum value of the expression \( E = \frac{x}{2-x} + \frac{y}{2-y} + \frac{z}{2-z} \) given that \( x + y + z = 1 \) and \( x, y, z > 0 \), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ E = \frac{x}{2-x} + \frac{y}{2-y} + \frac{z}{2-z} \]

### Step 2: Apply the Cauchy-Schwarz Inequality
Using the Cauchy-Schwarz inequality, we can write:
\[ \left( \frac{x}{2-x} + \frac{y}{2-y} + \frac{z}{2-z} \right) \left( (2-x) + (2-y) + (2-z) \right) \geq (x + y + z)^2 \]
Substituting \( x + y + z = 1 \):
\[ E \cdot (6 - (x + y + z)) \geq 1^2 \]
This simplifies to:
\[ E \cdot 5 \geq 1 \]
Thus,
\[ E \geq \frac{1}{5} \]

### Step 3: Find the Minimum Value
To find the minimum value, we need to check if this bound can be achieved. We can set \( x = y = z \) since they are symmetric in the expression. Given \( x + y + z = 1 \), we can set:
\[ x = y = z = \frac{1}{3} \]
Now substituting back into the expression:
\[ E = 3 \cdot \frac{\frac{1}{3}}{2 - \frac{1}{3}} = 3 \cdot \frac{\frac{1}{3}}{\frac{5}{3}} = 3 \cdot \frac{1}{5} = \frac{3}{5} \]

### Step 4: Conclusion
Thus, the minimum value of \( E \) is:
\[ \frac{3}{5} \]

### Final Answer
The minimum value of \( \frac{x}{2-x} + \frac{y}{2-y} + \frac{z}{2-z} \) is \( \frac{3}{5} \).

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