If `x+y+z=1`, then the minimum value of
`xy(x+y)^(2)+yz(y+z)^(2)+zx(z+x)^(2)` is , where x,y,z` inR^(+)`

Options

A. 4 xyz
B. 3 xyz
C. 2 xyz
D. 6 xyz

Explanation

To find the minimum value of the expression \( E = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2 \) given that \( x+y+z = 1 \) and \( x, y, z \in \mathbb{R}^+ \), we can follow these steps:

### Step 1: Substitute \( z \)
Since \( x + y + z = 1 \), we can express \( z \) in terms of \( x \) and \( y \):
\[ z = 1 - x - y \]

### Step 2: Rewrite the expression
Now we can rewrite the expression \( E \) in terms of \( x \) and \( y \):
\[ E = xy(x+y)^2 + y(1-x-y)(y+(1-x-y))^2 + (1-x-y)x(1-x-y+x)^2 \]

### Step 3: Simplify each term
1. For the first term:
\[ xy(x+y)^2 = xy(1-z)^2 = xy(1 - (1-x-y))^2 = xy(1 - z)^2 \]
2. For the second term:
\[ yz(y+z)^2 = y(1-x-y)(y + (1-x-y))^2 = y(1-x-y)(1-x)^2 \]
3. For the third term:
\[ zx(z+x)^2 = (1-x-y)x((1-x-y)+x)^2 = (1-x-y)x(1-y)^2 \]

### Step 4: Combine the terms
Combine all the terms together:
\[ E = xy(1-z)^2 + y(1-x-y)(1-x)^2 + (1-x-y)x(1-y)^2 \]

### Step 5: Analyze the expression
To minimize \( E \), we can apply the method of Lagrange multipliers or use inequalities. Here, we can apply the AM-GM inequality.

### Step 6: Apply AM-GM inequality
Using the AM-GM inequality:
\[ \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \]
Since \( x+y+z = 1 \):
\[ \frac{1}{3} \geq \sqrt[3]{xyz} \implies xyz \leq \frac{1}{27} \]

### Step 7: Substitute back into the expression
Now we can substitute \( xyz \) back into the expression for \( E \):
\[ E \geq 4xyz \]
Given that \( xyz \leq \frac{1}{27} \):
\[ E \geq 4 \cdot \frac{1}{27} = \frac{4}{27} \]

### Step 8: Conclusion
Thus, the minimum value of \( E \) occurs when \( x = y = z = \frac{1}{3} \):
\[ E = 4xyz = 4 \cdot \left(\frac{1}{3}\right)^3 = 4 \cdot \frac{1}{27} = \frac{4}{27} \]

### Final Answer
The minimum value of \( xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2 \) given \( x+y+z = 1 \) is \( \frac{4}{27} \).

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