Question

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

Answer 1

Calculation:

x and y are positive real numbers satisfying x + y = 102.

We know, Arithmetic Mean = (x + y)/2

Geometric Mean = √xy

Harmonic Mean =  \(\frac{2}{{\frac{1}{x} + \frac{1}{y}}}\)

Then, we know AM ≥ GM ≥ HM

⇒ (x + y)/2 ≥ √xy ≥ \(\frac{2}{{\frac{1}{x} + \frac{1}{y}}}\)

By (i) and (ii) term

⇒ (x + y)/2 ≥ √xy

⇒ 102/2 ≥ √xy

⇒ (51)² ≥ xy

⇒ 1/xy ≥ 1/2601

By (i) and (iii) term 

⇒ (x + y)/2 ≥ \(\frac{2}{{\frac{1}{x} + \frac{1}{y}}}\)

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

We have, \(2601\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{1}{y}} \right)\)

⇒ \(2601\;\left( {\;1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{{xy}}} \right)\)

⇒ 2601 × (1 + 2/51 + 1/2601)

⇒ 2704

∴ The minimum possible value of \(2601\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{1}{y}} \right)\) is 2704.