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

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Answer 1 · 2018-03-14T03:07:04

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

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

We want to find a $1/n$ such that $n + 1/n$ is minimized. We can call this sum a function $f(n) = n + 1/n$ .

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

$f'(n) = 1 -1/n^2$

$1 - 1/n^2 = 0$
$1 = 1/n^2$
$n^2 = 1$
$n = +- 1$

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

So the minimum sum obtainable is $f(1) = 1+ 1/1 = 2$

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