If x and y are two real numbers

Solution:

It is given that, x and y are two real numbers such that x > 0 and xy = 1.

Let $S=x+y$

Now, $x y=1 \Rightarrow y=\frac{1}{x}$

$\therefore S=x+y=x+\frac{1}{x}$

Differentiating both sides with respect to x, we get

$\frac{d S}{d x}=1-\frac{1}{x^{2}}$

For maxima or minima,

$\frac{d S}{d x}=0$

$\Rightarrow 1-\frac{1}{x^{2}}=0$

$\Rightarrow x^{2}=1$

$\Rightarrow x=1 \quad(x>0)$

Now,

$\frac{d^{2} S}{d x^{2}}=\frac{2}{x^{3}}$

At x = 1, we have

$\left(\frac{d^{2} S}{d x^{2}}\right)_{x=1}=\frac{2}{(1)^{3}}=2>0$

So, x = 1 is the point of local minimum

Thus, S is minimum when x = 1.

When $x=1, y=\frac{1}{x}=1$

$\therefore$ Minimum value of $S=x+y=1+1=2$

Thus, the minimum value of x + y is 2.

If x and y are two real numbers such that x > 0 and xy = 1. The the minimum value of x + y is ___2___.