Question

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

Answer 1

$AM≥GM≥HM$

$\frac{x+y}{2}≥\sqrt{xy}≥\frac{2}{\frac{1}{x}+\frac{1}{y}}$

Given $x+y=102$

$⇒ xy≤51^2 \;or\; \frac{1}{xy}≥\frac{1}{2601}$

$⇒\frac{ 1}{x}+\frac{1}{y}≥\frac{2}{51}$

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

= $2601\bigg(1+\frac{2}{51}+\frac{1}{2601}\bigg)$

= $2704$