x and y are two +ve numbers suchs that xy =1. Then the minimum value of x + y is

\frac{1}{4}

\frac{1}{2}

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x and y are two +ve numbers suchs that xy =1. Then the minimum value of x + y is

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Text solutionVerified

x y = 1

s = x + \frac{1}{x}

use AM & GM

\frac{x + \frac{1}{x}}{2} \geq (x \times \frac{1}{x})^{\frac{1}{2}}

x + \frac{1}{x} \geq 2

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Question Text	x and y are two +ve numbers suchs that xy =1. Then the minimum value of x + y is
Answer Type	Text solution:1
Upvotes	150