# How do you find a positive number such that the sum of the number and its reciprocal is as small as possible?

Mar 14, 2018

The smallest sum of a number $n$ and its reciprocal $\frac{1}{n}$ is $2$ which occurs when $n = 1$. Any other value of $n$ will produce a larger sum.

#### Explanation:

Let us consider a positive number $n$, making sure $n \setminus \ne 0$ so that we don't have an undefined reciprocal.

We want to find a $\frac{1}{n}$ such that $n + \frac{1}{n}$ is minimized. We can call this sum a function $f \left(n\right) = n + \frac{1}{n}$.

Now we take the derivative of $f \left(n\right)$ w.r.t. $n$ and set it equal to zero to obtain the minimum.

$f ' \left(n\right) = 1 - \frac{1}{n} ^ 2$

$1 - \frac{1}{n} ^ 2 = 0$
$1 = \frac{1}{n} ^ 2$
${n}^{2} = 1$
$n = \pm 1$

However, we reject the negative value as $n > 0$. Hence, $n = 1$.

So the minimum sum obtainable is $f \left(1\right) = 1 + \frac{1}{1} = 2$

Hence, the smallest sum of a number $n$ and its reciprocal $\frac{1}{n}$ is 2 when $n = 1$. Any other value of $n$ will produce a larger sum.