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Let $A,B,C,D$ and $E$ be five non-overlapping squares inside a unit square, of side lengths $a,b,c,d,e$, respectively. Prove that $a+b+c+d+e \le 2.$

(This problem is a continuation of my previous question about two squares in a box, which received an ingenious answer. If one asks for the maximum $M$ of the sum of the side lengths of $n$ nonoverlapping squares in a unit square, then the instances $n=3$ and $n=4$ yields $M=1.5$ and $M=2$ respectively. Proof: For $n$ squares with side lengths $a_1, ..., a_n$, the case $n=2$ implies $a_i+a_{i+1}\le1$ and thus, $M =\frac{1}{2} \sum\limits_{i=0}^na_i+a_{i+1}\le \frac{n}{2}$. To see that this upper bound can be attained for $n=3,4$, simply divide your square by a horizontal and a vertical line through its center into four equal squares and pick 3 or 4 of these squares. Therefore, $n=5$ is the next interesting case.)

EDIT: As this paper says, the problem has been solved by a certain D. J. Newman in a "personal communiction". Is anyone able to get his hands on this obscure paper?

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    A personal communication is not a paper; a personal communication is a personal communication. It might be something Newman told Graham in the elevator one day. There may be nothing to get one's hands on.2012-12-11

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