If a, b and c are distinct positive numbers, then the expression `(a + b - c)(b+ c- a)(c+ a -b)- abc` is:

Options

A. positive
B. negative
C. non-positive
D. non-negative

Explanation

To solve the expression \((a + b - c)(b + c - a)(c + a - b) - abc\) for distinct positive numbers \(a\), \(b\), and \(c\), we can follow these steps:

### Step 1: Analyze the terms
We start with the expression:
\[ E = (a + b - c)(b + c - a)(c + a - b) - abc \]
We need to determine whether \(E\) is positive, negative, or zero.

### Step 2: Apply the AM-GM Inequality
We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to each pair of terms.

1. For the first two terms:
\[ \frac{(a + b - c) + (b + c - a)}{2} \geq \sqrt{(a + b - c)(b + c - a)} \]
Simplifying the left side gives:
\[ \frac{(a + b + b + c - c - a)}{2} = \frac{2b}{2} = b \]
Thus, we have:
\[ b \geq \sqrt{(a + b - c)(b + c - a)} \]

2. For the next pair:
\[ \frac{(b + c - a) + (c + a - b)}{2} \geq \sqrt{(b + c - a)(c + a - b)} \]
This simplifies to:
\[ \frac{2c}{2} = c \]
So:
\[ c \geq \sqrt{(b + c - a)(c + a - b)} \]

3. For the last pair:
\[ \frac{(c + a - b) + (a + b - c)}{2} \geq \sqrt{(c + a - b)(a + b - c)} \]
This simplifies to:
\[ \frac{2a}{2} = a \]
Hence:
\[ a \geq \sqrt{(c + a - b)(a + b - c)} \]

### Step 3: Multiply the inequalities
Now, we multiply the three inequalities:
\[ abc \geq \sqrt{(a + b - c)(b + c - a)} \cdot \sqrt{(b + c - a)(c + a - b)} \cdot \sqrt{(c + a - b)(a + b - c)} \]
This implies:
\[ abc \geq \sqrt{(a + b - c)(b + c - a)(c + a - b)}^2 \]
Thus, we have:
\[ abc \geq (a + b - c)(b + c - a)(c + a - b) \]

### Step 4: Rearranging the expression
Rearranging gives us:
\[ (a + b - c)(b + c - a)(c + a - b) - abc \leq 0 \]
This means:
\[ E \leq 0 \]

### Step 5: Conclusion
Since \(a\), \(b\), and \(c\) are distinct positive numbers, the expression cannot equal zero. Therefore, we conclude that:
\[ E < 0 \]

### Final Answer
The expression \((a + b - c)(b + c - a)(c + a - b) - abc\) is negative.

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