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Minimum of x^2 + y^2 Given xy = 1 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

x \cdot y = 1

, then the minimum value of

x^{2}+y^{2}

1

\sqrt2

2

correct

4

Solution

Step 1: Use the constraint to remove one variable.

xy = 1 \implies y = \frac{1}{x}, \qquad x \ne 0 .

Step 2: Substitute.

x^{2}+y^{2} = x^{2} + \frac{1}{x^{2}} .

Step 3: Apply AM

\ge

GM to the two positive quantities

x^{2}

and

\dfrac{1}{x^{2}}

\frac{x^{2}+\dfrac{1}{x^{2}}}{2} \ \ge\ \sqrt{x^{2}\cdot\frac{1}{x^{2}}} = 1 .

Step 4: Multiply by

2

x^{2}+y^{2} \ \ge\ 2 .

Step 5: Check the bound is reached. Equality in AM

\ge

GM needs

x^{2} = \dfrac{1}{x^{2}}

, i.e.

x^{4} = 1

, i.e.

x = \pm1

. Then

y = \pm1

with

xy = 1

and

x^{2}+y^{2} = 1+1 = 2 .

Answer: (3).

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