Prove that If any A B C are distinct positive numbers , then the expression (b+c-a)(c+a-b)(a+b-c)-abc is negative

Answer

To prove that the expression \((b+c-a)(c+a-b)(a+b-c) - abc < 0\) for distinct positive numbers \(a\), \(b\), and \(c\), we can follow these steps:

### Step 1: Define the variables
Let:
- \(x = b + c - a\)
- \(y = c + a - b\)
- \(z = a + b - c\)

### Step 2: Analyze the conditions
Since \(a\), \(b\), and \(c\) are distinct positive numbers, we know that:
- \(x > 0\)
- \(y > 0\)
- \(z > 0\)

This is because each of these expressions represents the sum of two positive numbers minus a distinct positive number.

### Step 3: Apply the AM-GM inequality
By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \]
This implies:
\[ x + y + z \geq 3\sqrt[3]{xyz} \]

### Step 4: Calculate \(x + y + z\)
Now, we can calculate \(x + y + z\):
\[ x + y + z = (b+c-a) + (c+a-b) + (a+b-c) = (b+c-a+c+a-b+a+b-c) = a + b + c \]

### Step 5: Substitute back into the inequality
Thus, we can rewrite the inequality from AM-GM:
\[ \frac{a + b + c}{3} \geq \sqrt[3]{xyz} \]

### Step 6: Calculate \(xyz\)
Now, we need to calculate \(xyz\):
\[ xyz = (b+c-a)(c+a-b)(a+b-c) \]

### Step 7: Compare \(xyz\) with \(abc\)
We want to show that:
\[ xyz - abc < 0 \]
This means we need to show that \(xyz < abc\).

### Step 8: Conclude the proof
From the AM-GM inequality, we can conclude that:
\[ xyz < \left(\frac{x + y + z}{3}\right)^3 = \left(\frac{a + b + c}{3}\right)^3 \]
And since \(a\), \(b\), and \(c\) are distinct positive numbers, we have:
\[ \left(\frac{a + b + c}{3}\right)^3 < abc \]
Thus, we can conclude that:
\[ (b+c-a)(c+a-b)(a+b-c) - abc < 0 \]

### Final Result
Therefore, we have proved that:
\[ (b+c-a)(c+a-b)(a+b-c) - abc < 0 \]