Question 108

If $$x$$ and $$y$$ are positive real numbers satisfying $$x + y = 52$$ , then the minimum possible value of $$91(1 + \frac{1}{x})(1 + \frac{1}{y})$$ is:

Solution

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Lets expand the given equation.

$$91\left(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\right)$$

$$91\left(1+\frac{x+y}{xy}+\frac{1}{xy}\right)$$

x+y is constant, 91 is constant.

To get the minimum value we have to minimixe the xy term.

That happens when both are equal.

So x=26, y=26.

$$91\left(\left(1+\frac{1}{26}\right)\left(1+\frac{1}{26}\right)\right)$$

On simiplifing we get $$\frac{5103}{52}$$

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