You are true! Following problem is difficult than above problem. Let $n$ be positive integer. If $x_{1},x_{2},...,x_{n}$ are positive integers with $\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}<1$, find the maximum possible value of $\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}.$

This is the first time in the pre-olympiad forum.

To rigorize cincodemayo's solution, we only need to note that if we don't have a $\frac{1}{2}$ then we must have all three fractions at most $\frac{1}{3}$ and one which is at most $\frac{1}{4}$, so the maximal possible sum is $\frac{1}{3}+\frac{1}{3}+\frac{1}{4}= \frac{11}{12}$. 1=2's solution extends it with a greedy approach, which certainly seems like it should always give the . The question is, can you prove to me that there is never a better solution? Incidentally, here is the entry for the sequence 1=2 suggests in the OEIS.

for sake of contradiction, suppose that there exists a triple (a,b,c) with $a\ge b\ge c$ such that $1>\sum\frac{1}{a}>\frac{41}{42}$ we have that $\frac{41}{42}< \sum\frac{1}{a}\le \frac{3}{c}$, hence, $c\le \frac{126}{41}=3.07...$, then cases... $c=3$, we have $\frac{1}{a}+\frac{1}{b}>\frac{9}{14}\implies \frac{2}{b}>\frac{9}{14}\implies b<\frac{28}{9}=3.11...$, but $b\ge c$, so $b=3$, and we have $\frac{1}{a}>\frac{13}{42}\implies a<\frac{42}{13}=3.23...$ but $a\ge b=3$, so $a=3$, giving our sum to be $1$, contradiction else, c=2, and follow sam's logic [which i have not looked at, but i presume to be correct]

N.T.TUAN wrote: You are true! Following problem is difficult than above problem. Let $n$ be positive integer. If $x_{1},x_{2},...,x_{n}$ are positive integers with $\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}<1$, find the maximum possible value of $\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}.$

$ml+mk+kl\ge mlk-1$, $\frac{1}{m}+\frac{1}{l}+\frac{1}{k}\ge 1-\frac{1}{mlk}$, so we have to maximize $mlk$. Suppose $m \ge l \ge k$. If $3 \le k \le m \le g$ we have contradiction(because $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ exceeds $1$), so suppose $k=2$. Now, if $l\ge 4$ we obtain $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}\le\frac{5}{6}$, so take $l=3$, and of course $m=7$ and we obtain $\frac{41}{42}>\frac{5}{6}$.