If a,b,c are distinct positive real numbers, then

Options

A. `a^(2)+b^(2)+c^(2)gtab+bc+ca`
B. `a^(2)+b^(2)+c^(2)ltab+bc+ca`
C. `a^(2)+b^(2)+c^(2)geab+bc+ca`
D. `a^(2)+b^(2)+c^(2)leab+bc+ca`

Explanation

To solve the problem step by step, we will use the properties of inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step 1: Understand the problem
We are given three distinct positive real numbers \( a, b, c \) and we need to show that \( a^2 + b^2 + c^2 > ab + bc + ca \).

### Step 2: Apply the AM-GM Inequality
According to the AM-GM inequality, for any two distinct positive real numbers \( x_1 \) and \( x_2 \):
\[ \frac{x_1 + x_2}{2} \geq \sqrt{x_1 x_2} \]
This implies:
\[ x_1 + x_2 \geq 2\sqrt{x_1 x_2} \]

### Step 3: Apply AM-GM to pairs of squares
1. **For \( a^2 \) and \( b^2 \)**:
\[ \frac{a^2 + b^2}{2} \geq \sqrt{a^2 b^2} \implies a^2 + b^2 \geq 2ab \]

2. **For \( b^2 \) and \( c^2 \)**:
\[ \frac{b^2 + c^2}{2} \geq \sqrt{b^2 c^2} \implies b^2 + c^2 \geq 2bc \]

3. **For \( c^2 \) and \( a^2 \)**:
\[ \frac{c^2 + a^2}{2} \geq \sqrt{c^2 a^2} \implies c^2 + a^2 \geq 2ca \]

### Step 4: Add the inequalities
Now, we add the three inequalities obtained:
\[ (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) \geq 2ab + 2bc + 2ca \]
This simplifies to:
\[ 2(a^2 + b^2 + c^2) \geq 2(ab + bc + ca) \]

### Step 5: Divide by 2
Dividing both sides by 2 gives:
\[ a^2 + b^2 + c^2 \geq ab + bc + ca \]

### Step 6: Analyze the distinctness
Since \( a, b, c \) are distinct positive real numbers, the equality \( a^2 + b^2 + c^2 = ab + bc + ca \) cannot hold. Therefore, we conclude:
\[ a^2 + b^2 + c^2 > ab + bc + ca \]

### Conclusion
Thus, we have shown that:
\[ a^2 + b^2 + c^2 > ab + bc + ca \]
This confirms the correct option is the third one.

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