Bugi wrote: Let $ n\ge 3$ be a natural number. A set of real numbers $ \{x_1,x_2,\ldots,x_n\}$ is called summable if $ \sum_{i = 1}^n \frac {1}{x_i} = 1$. Prove that for every $ n\ge 3$ there always exists a summable set which consists of $ n$ elements such that the biggest element is: a) bigger than $ 2^{2n - 2}$ b) smaller than $ n^2$ What if $ 2^{2n-2} > n^2$?

The cases are 2 different, you must prove that each one of them holds, not both of them in one.

Bugi wrote: Let $ n\ge 3$ be a natural number. A set of real numbers $ \{x_1,x_2,\ldots,x_n\}$ is called summable if $ \sum_{i = 1}^n \frac {1}{x_i} = 1$. Prove that for every $ n\ge 3$ there always exists a summable set which consists of $ n$ elements such that the biggest element is: a) bigger than $ 2^{2n - 2}$ b) smaller than $ n^2$ a) Induction on $ n$ but i think it should be bigger than $ 2^{2(n-2)}$ For the case of $ n=3$, $ S_3=\{2,3,6\}$ works Notice for the inductive step that if $ S_i$ is summable then $ \{2\} \cup 2 \cdot S_i$ is also summable .

mszew wrote: Bugi wrote: Let $ n\ge 3$ be a natural number. A set of real numbers $ \{x_1,x_2,\ldots,x_n\}$ is called summable if $ \sum_{i = 1}^n \frac {1}{x_i} = 1$. Prove that for every $ n\ge 3$ there always exists a summable set which consists of $ n$ elements such that the biggest element is: a) bigger than $ 2^{2n - 2}$ b) smaller than $ n^2$ a) Induction on $ n$ but i think it should be bigger than $ 2^{2(n - 2)}$ For the case of $ n = 3$, $ S_3 = \{2,3,6\}$ works Notice for the inductive step that if $ S_i$ is summable then $ \{2\} \cup 2 \cdot S_i$ is also summable . Then $ S_4 = \{2,4,6,12\}$ But $ 12 < 2^{2(4 - 2)} = 2^4 = 16$. Bugi: If $ x_1,...,x_n$ is real numbers, then it is trivial. Shouldn't it be $ x_1,..,x_n$ are whole numbers? For $ b$ we just set $ x_1 = x_2 = ... = x_n = n < n^2$.

Sorry for the mistake, I believe that all $ x_i$ should be different. Notice that $ \frac {1}{n} = \frac {1}{n + 1} + \frac {1}{n(n + 1)}$ then the inductive step will be $ \{x_1,\cdots,x_{n - 1},x_n\} \rightarrow \{x_1,\cdots,x_{n - 1},x_n + 1,x_n \cdot (x_n + 1)\}$ For example $ \{2,3,6\} \rightarrow \{2,3,7,42\} \rightarrow \{2,3,7,43,1806\}$

I don't know, I just posted this: http://www.mathlinks.ro/viewtopic.php?t=212129 Look at that topic

I've got the shortlist form orl and there is this problem a bit different: Problem A3. Let $ n \geq 3$ be an integer. A set of positive integers $ x_1,x_2,...,x_n$ is called summable if $ \frac {1}{x_1} + \frac {1}{x_2} + ... + \frac {1}{x_n} = 1$. (i) Prove that for every $ n \geq 3$ there exists a summable set of $ n$ integers having the value of the largest element greater than $ 2^{2^{n - 2}}$; (ii) Prove that for every $ n\geq 3$ there exists a summable set of $ n$ distinct integers having the value of the largest element less than $ n^2$.

what's the original problem?!it's wrong for $n=3,4$!

I have an approach through inequality. We define the set {x1,x2,.....xn} such that the sequence is strictly increasing. Thua, without loss of generality,suppose xn is greatest. Now we have to chose xn twice, once such that it is bigger than 2^(2n-2) and when it is smaller than n^2. Now, Working Backward, suppose the set is summable. Then, x1+x2+x3+.....xn >= n^2. [AM>= HM]. Then we can chose all of x_i to be less than n^2 but greater than n, and we see thus for such a selection, the inequality holds. Thus, the set is trivially summable for such a selection.(2nd part proved) Again, we are done with the first part if we can select xn> 2^(2n-2). Suppose all x_i greater than 2^(2n-2). Then also, we see that the inequality reduces to n. 2^(2n-2) > n^2 , which is obvious for n>=3.