If `xy=1` , then minimum value of `x ^(2) + y^(2)` is :

Options

A. 1
B. 2
C. `sqrt2`
D. 4

Answer

To find the minimum value of \( x^2 + y^2 \) given that \( xy = 1 \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Start with the Given Condition**:
We are given that \( xy = 1 \).

2. **Apply the AM-GM Inequality**:
According to the AM-GM inequality, for any non-negative real numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
In our case, let \( a = x^2 \) and \( b = y^2 \). Therefore:
\[ \frac{x^2 + y^2}{2} \geq \sqrt{x^2 y^2} \]

3. **Simplify the Right Side**:
Since \( xy = 1 \), we have:
\[ \sqrt{x^2 y^2} = \sqrt{(xy)^2} = \sqrt{1^2} = 1 \]
Thus, the inequality becomes:
\[ \frac{x^2 + y^2}{2} \geq 1 \]

4. **Multiply Both Sides by 2**:
To eliminate the fraction, multiply both sides by 2:
\[ x^2 + y^2 \geq 2 \]

5. **Conclusion**:
The minimum value of \( x^2 + y^2 \) is \( 2 \). This minimum occurs when \( x = y \). Given \( xy = 1 \), if \( x = y \), then \( x^2 = 1 \) and thus \( x = y = 1 \).

### Final Answer:
The minimum value of \( x^2 + y^2 \) is \( 2 \).

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