Application of DerivativesmediumFree

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

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Question
If x⋅y=1x \cdot y = 1, then the minimum value of x2+y2x^{2}+y^{2} is
A11
B2\sqrt2
C22correct
D44
Solution
Step 1: Use the constraint to remove one variable.
xy=1  ⟹  y=1x,x≠0.xy = 1 \implies y = \frac{1}{x}, \qquad x \ne 0 .
 Step 2: Substitute.
x2+y2=x2+1x2.x^{2}+y^{2} = x^{2} + \frac{1}{x^{2}} .
 Step 3: Apply AM ≥\ge GM to the two positive quantities x2x^{2} and 1x2\dfrac{1}{x^{2}}.
x2+1x22 ≥ x2⋅1x2=1.\frac{x^{2}+\dfrac{1}{x^{2}}}{2} \ \ge\ \sqrt{x^{2}\cdot\frac{1}{x^{2}}} = 1 .
 Step 4: Multiply by 22.
x2+y2 ≥ 2.x^{2}+y^{2} \ \ge\ 2 .
 Step 5: Check the bound is reached. Equality in AM ≥\ge GM needs x2=1x2x^{2} = \dfrac{1}{x^{2}}, i.e. x4=1x^{4} = 1, i.e. x=±1x = \pm1. Then y=±1y = \pm1 with xy=1xy = 1 and
x2+y2=1+1=2.x^{2}+y^{2} = 1+1 = 2 .
 Answer: (3).
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