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

The correct answer is 2704.

We are given that x and y are positive real numbers with x + y = 102, and we need to find the minimum value of:

$$2601(1+\frac{1}{x})(1+\frac{1}{y})$$

Step 1: Expand the expression inside.

First, let's expand the product $(1+\frac{1}{x})(1+\frac{1}{y})$:

$$=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}$$

Combining the middle two fractions over a common denominator xy:

$$=1+\frac{x+y}{xy}+\frac{1}{xy}=1+\frac{x+y+1}{xy}$$

Step 2: Substitute the constraint x + y = 102.

Replacing x + y with 102:

$$=1+\frac{102+1}{xy}=1+\frac{103}{xy}$$

So the full expression becomes:

$$2601(1+\frac{103}{xy})=2601+\frac{2601\times 103}{xy}$$

Step 3: Determine what minimizes the expression.

Since 2601 is a constant and 103 is positive, the term $\frac{2601\times 103}{xy}$ decreases as xy increases. Therefore, to minimize the entire expression, we need to maximize the product xy.

Step 4: Maximize xy using the AM-GM Inequality.

The AM-GM inequality states that for positive real numbers:

$$\frac{x+y}{2}\ge \sqrt{xy}$$

Squaring both sides:

$$xy\le {\left(\frac{x+y}{2}\right)}^2={\left(\frac{102}{2}\right)}^2={51}^2=2601$$

Equality holds when x = y = 51. So the maximum value of xy is 2601.

Step 5: Compute the minimum value.

Substituting xy = 2601 into the expression:

$$2601+\frac{2601\times 103}{2601}=2601+103=2704$$

Elegant Observation: Notice that $2601={51}^2$ and $2704={52}^2$. The minimum value is the perfect square of (51 + 1) = 52, which is a beautiful result!

Verification: At x = y = 51:
$$2601(1+\frac{1}{51})(1+\frac{1}{51})=2601\times {\left(\frac{52}{51}\right)}^2=2601\times \frac{2704}{2601}=2704✓$$

Therefore, the minimum possible value is 2704.