Minimum Value of Expression

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601 $(1 + \frac{1}{x}) (1 + \frac{1}{y})$ is

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CAT 2020 Slot-2

Inequalities

Medium

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601 $(1 + \frac{1}{x}) (1 + \frac{1}{y})$ is

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