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

We know that, $\boxed{\text{AM} \geq \text{GM} \geq \text{HM}}$

$ \Rightarrow \frac{x+y}{2} \geq \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $

$ \Rightarrow \frac{102}{2} \geq \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} \quad \longrightarrow (1)$

Take the first two terms, from equation $(1).$

$ 51 \geq \sqrt{xy} $

$ \Rightarrow xy \leq 51^{2} $

$ \Rightarrow \boxed{xy \leq 2601} $

Taking the last two terms, from equation $(1).$

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

$ \Rightarrow 51 \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $

$ \Rightarrow \boxed{\frac{1}{x} + \frac{1}{y} \geq \frac{2}{51}} $

Now, the minimum possible 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) $

$\qquad  = 2601 \left( 1 + \frac{2}{51} + \frac{1}{2601} \right)  = 2601 \left( \frac{2601 + 102 + 1}{2601} \right)  = 2704 $

$\therefore$ The minimum possible value is $2704.$

Correct Answer$: 2704$
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