## if $x + y = 4$ for positive integers x and y , then the smallest possible value of $\frac{1}{x} + \frac{1}{y}$

May 13, 2018

1

#### Explanation:

There are only two possible combinations of positive integer $x$ and $y$. These are $2$ and $2$, and $3$ and $1$. Luckily, this is not too exhaustive to test both of these.

Let $x = 2$ and $y = 2$:

$\frac{1}{2} + \frac{1}{2} = 1$

Let $x = 3$ and $y = 1$

$\frac{1}{3} + \frac{1}{1} = \frac{4}{3}$

Therefore, since $1 < \frac{4}{3}$, the minimum value is 1

May 13, 2018

Isn't it 1

#### Explanation:

$x + y = 4$
$\implies x = 4 - y$

Substituting in the equation:
$= \frac{1}{4 - y} + \frac{1}{y}$

LCM of denominators is $y \left(4 - y\right)$
$= \frac{y}{\left(4 - y\right) y} + \frac{4 - y}{y \left(4 - y\right)}$

Combining the denominators:
$= \frac{y \left(4 - y\right)}{y \left(4 - y\right)}$

$= 1$

May 13, 2018

Given: $x + y = 4 , x > 0 , y > 0$

Minimize:

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

Substitute $y = 4 - x$:

$\frac{1}{x} + \frac{1}{4 - x}$

Compute the first derivative with respect to x:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x} ^ 2 + \frac{1}{4 - x} ^ 2$

To find local extrema we set the first derivative equal to 0:

$0 = - \frac{1}{x} ^ 2 + \frac{1}{4 - x} ^ 2$

Multiply both sides by $- {x}^{2} {\left(4 - x\right)}^{2}$:

$0 = {\left(4 - x\right)}^{2} - {x}^{2}$

Expand the square:

$0 = {x}^{2} - 8 x + 16 - {x}^{2}$

$0 = - 8 x + 16$

$x = 2$

Find the corresponding value of y:

$y = 4 - 2$

$y = 2$

The value is:

$\frac{1}{2} + \frac{1}{2} = 1$

We could perform the second derivative test but we know that this is a minimum because it is obvious that the maximum occurs when either $x$ or $y$ approaches 0.