David Johnson 1973 maximum sum of subdivisions of 1 On "Johnson, D. S. (1973). Near-optimal bin packing algorithms (Doctoral dissertation, Massachusetts Institute of Technology)", David Johson argues on page 65 that, given positive integers $x_{1},\dots,x_{n}$  where $n\geq3$ , such that $\sum_{i=1}^{n}{1/x_{i}}=1$ (which he calls a "subdivision of 1"), then the maximum value of $\sum_{i=1}^{n}{1/(x_{i}-1)}$  is 17/10.
He shows that this bound is achieved with $n=3$ , $x_1=2,x_2=3,x_3=6$, fair enough. But omits the general proof that no other set of integers can do better, saying only that it is "easily derived". How can we prove this ?
 A: Looking back, I could agree that it is "simple", but it took me some weeks to finally reach this conclusion:
Let us order the sequence of integers so that $x_1\leq x_2\leq \dots \leq x_n$ . Notice that $x_1 > 1$ .

*

*If $x_1=2$ , $x_2$  cannot be 2, otherwise $1/x_1+1/x_2=1$ and there could not be other integers  $x_3,\dots x_n$ .

1.1) If $x_2=3$ , then $\sum_{i=3}^{n}{1/x_i}=1/6$ , therefore $x_i\geq 6$ for all $i\geq3$ . But notice that
$$
\begin{align}
\sum_{i=3}^{n}{\frac{1}{(x_i-1)}} & = \sum_{i=3}^{n}{\frac{1}{x_i}} + \sum_{i=3}^{n}{\frac{1}{x_i(x_i-1)}} \\
 & = \frac{1}{6} + \sum_{i=3}^{n}{\frac{1}{x_i(x_i-1)}} \\
 & \leq \frac{1}{6} + \frac{1}{6}\sum_{i=3}^{n}{\frac{1}{(x_i-1)}} \\
\implies \sum_{i=3}^{n}{\frac{1}{(x_i-1)}} \leq \frac{1}{5}
\end{align}
$$
So $\sum_{i=1}^{n}{1/(x_i-1)} \leq 1 + 1/2 + 1/5 = 17/10$ .
1.2) If $x_2 \geq 4$ , then $x_i\geq4$  for all $i\geq3$ and, similarly:
$$
\begin{align}
\sum_{i=2}^{n}{\frac{1}{(x_i-1)}} & = \sum_{i=2}^{n}{\frac{1}{x_i}} + \sum_{i=2}^{n}{\frac{1}{x_i(x_i-1)}} \\
 & = \frac{1}{2} + \sum_{i=2}^{n}{\frac{1}{x_i(x_i-1)}} \\
 & \leq \frac{1}{2} + \frac{1}{4}\sum_{i=2}^{n}{\frac{1}{(x_i-1)}} \\
\implies \sum_{i=2}^{n}{\frac{1}{(x_i-1)}} \leq \frac{2}{3}
\end{align}
$$
So $\sum_{i=1}^{n}{1/(x_i-1)} \leq 1 + 2/3 < 17/10$ .


*If $x_1\geq3$ , then $x_i\geq3$ for all $i\geq2$ , thus:

$$
\begin{align}
\sum_{i=1}^{n}{\frac{1}{(x_i-1)}} & = \sum_{i=1}^{n}{\frac{1}{x_i}} + \sum_{i=1}^{n}{\frac{1}{x_i(x_i-1)}} \\
 & = 1 + \sum_{i=1}^{n}{\frac{1}{x_i(x_i-1)}} \\
 & \leq 1 + \frac{1}{3}\sum_{i=1}^{n}{\frac{1}{(x_i-1)}} \\
\implies \sum_{i=1}^{n}{\frac{1}{(x_i-1)}} \leq \frac{3}{2} < 17/10
\end{align}
$$
Thus the claim is proved.
