Question
If x and y are positive real numbers satisfying x+y=102, then the minimum possible value of \(2601\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{1}{y}} \right)\) is
Option: 2704
Solution:
Solution:
\(AM \ge GM \ge HM\)
\(\frac{{x + y}}{2} \ge \sqrt {xy} \ge \frac{2}{{\frac{1}{x} + \frac{1}{y}}}\)
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
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