prove the inequality where concerned number are all positive `a^2+b^2+c^2>ab+bc+ca`

Answer

To prove the inequality \( a^2 + b^2 + c^2 > ab + bc + ca \) for positive numbers \( a, b, c \), we can follow these steps:

### Step 1: Rearranging the Inequality
We start with the inequality we want to prove:
\[ a^2 + b^2 + c^2 > ab + bc + ca \]
We can rearrange this to:
\[ a^2 + b^2 + c^2 - ab - bc - ca > 0 \]

### Step 2: Grouping Terms
Next, we can group the terms in pairs:
\[ = \frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) \]
This step uses the identity:
\[ x^2 + y^2 = (x-y)^2 + 2xy \]
to express the left-hand side in terms of squared differences.

### Step 3: Analyzing the Expression
Since \( a, b, c \) are all positive numbers, the squared terms \( (a-b)^2, (b-c)^2, (c-a)^2 \) are all non-negative. Therefore, we have:
\[ (a-b)^2 \geq 0, \quad (b-c)^2 \geq 0, \quad (c-a)^2 \geq 0 \]
Adding these non-negative terms gives:
\[ (a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0 \]

### Step 4: Conclusion
Thus, we conclude that:
\[ \frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) \geq 0 \]
This implies:
\[ a^2 + b^2 + c^2 - ab - bc - ca > 0 \]
Therefore, we have proved that:
\[ a^2 + b^2 + c^2 > ab + bc + ca \]