Question:medium

If x and y are positive real numbers satisfying \(x+y=102\), then the minimum possible value of \(2601(1+\frac 1x)(1+\frac 1y)\) is
[This Question was asked as TITA]

Updated On: Jan 15, 2026
  • 2432
  • 2807
  • 2704
  • 2605
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The Correct Option is C

Solution and Explanation

The inequality \( AM \geq GM \geq HM \) is utilized, which implies \( \frac{x + y}{2} \geq \sqrt{xy} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}} \).

Given \( x + y = 102 \), applying \( AM \geq GM \) yields \( \sqrt{xy} \leq \frac{102}{2} = 51 \). Consequently, \( xy \leq 51^2 = 2601 \), and \( \frac{1}{xy} \geq \frac{1}{2601} \).

The \( HM \) inequality gives \( \frac{1}{x} + \frac{1}{y} \geq \frac{2}{51} \).

The expression \( 2601 \left(1 + \frac{1}{x} \right)\left(1 + \frac{1}{y} \right) \) is expanded as \( 2601 \left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \right) = 2601 \left( 1 + \left( \frac{1}{x} + \frac{1}{y} \right) + \frac{1}{xy} \right) \).

Substituting the minimum values determined previously, the expression becomes \( 2601 \left( 1 + \frac{2}{51} + \frac{1}{2601} \right) \).

Simplification yields \( 2601 \left( 1 + \frac{2}{51} + \frac{1}{2601} \right) = 2601 \times \left( \frac{2601 + 102 + 1}{2601} \right) = 2601 \times \frac{2704}{2601} = 2704 \).

∴ The minimum value is \( \boxed{2704} \).
Correct option is (C): 2704

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