CAT 2020 Quant Question [Slot 2] with Solution 06

\(AM \ge GM \ge HM\)

\(\frac{{x + y}}{2} \ge \sqrt {xy}  \ge \frac{2}{{\frac{1}{x} + \frac{1}{y}}}\)

Given \(x + y = 102\)

\( \Rightarrow xy \le {51^2}\) or \(\frac{1}{{xy}} \ge \frac{1}{{2601}}\)

\( \Rightarrow \frac{1}{x} + \frac{1}{y} \ge \frac{2}{{51}}\)

The minimum value of \(2601\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{1}{y}} \right)\) \( = (2601)\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{{xy}}} \right)\)

\( = 2601\left( {1 + \frac{2}{{51}} + \frac{1}{{2601}}} \right)\)

=2704