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

We need to find the minimum value of $2601(1+\tfrac{1}{x})(1+\tfrac{1}{y})$ given that $x + y = 102$.

Let us start by expanding the expression to better understand what we're working with.

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$2601(1+\tfrac{1}{x})(1+\tfrac{1}{y}) = 2601(1 + \tfrac{1}{x} + \tfrac{1}{y} + \tfrac{1}{xy})$

Since $x + y = 102$, we can simplify $\tfrac{1}{x} + \tfrac{1}{y}$:

$\tfrac{1}{x} + \tfrac{1}{y} = \tfrac{y + x}{xy} = \tfrac{102}{xy}$

So our expression becomes:

$2601(1 + \tfrac{102}{xy} + \tfrac{1}{xy}) = 2601(1 + \tfrac{103}{xy})$

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To minimize $2601(1 + \tfrac{103}{xy})$, we need to maximize $xy$ (since $\tfrac{103}{xy}$ gets smaller as $xy$ gets larger).

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The AM-GM Inequality states: For positive numbers $x$ and $y$, $\tfrac{x+y}{2} \geq \sqrt{xy}$

This means: $\tfrac{102}{2} \geq \sqrt{xy}$

Therefore: $51 \geq \sqrt{xy}$

$xy \leq 51^2 = 2601$

The maximum value of $xy$ is $2601$, and this happens when $x = y = 51$ (equality in AM-GM occurs when the numbers are equal).

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When $x = y = 51$:

$xy = 51 \times 51 = 2601$

Using the original form:

$2601(1+\tfrac{1}{51})(1+\tfrac{1}{51}) = 2601(\tfrac{52}{51})^2$

$= 2601 \times \tfrac{52^2}{51^2}$

$= 2601 \times \tfrac{2704}{2601}$

$= 2704$

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We used the fact that for a fixed sum, the product is maximized when the numbers are equal. This is a consequence of the AM-GM inequality, which is a powerful tool for optimization problems. The symmetry occurs because the optimal solution has $x = y$.

Therefore, the minimum possible value is $2704$.