If a,b,c are real numbers such that `a^(2)+b^(2)+c^(2)=1` then `ab+bc+cagt-1/2`.

Answer

To prove that if \( a, b, c \) are real numbers such that \( a^2 + b^2 + c^2 = 1 \), then \( ab + bc + ca \geq -\frac{1}{2} \), we can follow these steps:

### Step 1: Start with the given condition
We know that:
\[ a^2 + b^2 + c^2 = 1 \]

### Step 2: Use the square of a sum
Consider the expression \( (a + b + c)^2 \). Expanding this, we get:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \]

### Step 3: Substitute the known value
Since we know \( a^2 + b^2 + c^2 = 1 \), we can substitute this into our equation:
\[ (a + b + c)^2 = 1 + 2(ab + bc + ca) \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ 2(ab + bc + ca) = (a + b + c)^2 - 1 \]

### Step 5: Analyze the right-hand side
Since the square of any real number is non-negative, we have:
\[ (a + b + c)^2 \geq 0 \]
Thus:
\[ (a + b + c)^2 - 1 \geq -1 \]

### Step 6: Substitute back
This leads to:
\[ 2(ab + bc + ca) \geq -1 \]

### Step 7: Divide by 2
Dividing both sides by 2 gives:
\[ ab + bc + ca \geq -\frac{1}{2} \]

### Conclusion
We have shown that:
\[ ab + bc + ca \geq -\frac{1}{2} \]
This completes the proof.

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