Decompose a positive number into the sum of 5 numbers in such a way that the sum of its inverse is maximal.

Question

I have to decompose a positive number into the sum of 5 numbers in such a way that the sum of its inverse is maximal. I understand that is optimization, but I don't know how to do it with five variables.

Answer 1

Let $N$ be any positive number and let $a,b,c,d,e \in \mathbb{R}$ be such that $$ N = a + b + c + d + e. $$ Without any restrictions on what these $a,b,c,d,e$ are, then you can choose your numbers such that the sum of the reciprocals can take any value you want.

For the sake of example, let $b = N - a - 3$ and $c,d,e = 1$. Then $$ \begin{align} \sup_{a \in \mathbb{R}}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e}\right) &= \sup_{a \in \mathbb{R}}\left(\frac{1}{a} + \frac{1}{N-a-3}\right) +3 \\ &= \lim_{\epsilon\to 0}\left(\frac{1}{\epsilon} + \frac{1}{N - \epsilon - 3}\right) + 3 \\ &= + \infty \end{align} $$

Answer 2

Your question as I understand it is:

Given $\ \alpha > 0,\ $ what is the maximum, or more generally the supremum of $$S = \ \left\{ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\ :a+b+c+d+e = \alpha\right\}\quad ? $$

There is no maximal number or supremum of $S.\ $ To see this, suppose by way of contradiction such a maximal number or supremum, $M\in\mathbb{R},$ of $S,$ exists.

We get to choose $\ a,b,c,d,e.\ $ Choose $\ a = \min\left\{\frac{1}{2M}, \frac{\alpha}{2}\right\}.\ $ Then $\ a < \alpha\ $ and $\ \frac{1}{a} > M.$

Then just choose $\ b=c=d=e=\frac{\alpha-a}{4}\ $ so that $\ a+b+c+d+e = \alpha\ $. Note also that $\ b,\ c,\ d,\ e\ $ are all $\ > 0,\ $ and therefore $\ \frac{1}{b},\ \frac{1}{c},\ \frac{1}{d},\ \frac{1}{e}\ $ are all $\ > 0,\ $ and so $\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e} > \frac{1}{a} > M,\ $ contradicting the definition of $\ M,\ $ as desired.