Find the smallest real number $M$ with the following property: Given nine nonnegative real numbers with sum $1$, it is possible to arrange them in the cells of a $3 \times 3$ square so that the product of each row or column is at most $M$. Evan O' Dorney.
Problem
Source: ELMO Shortlist 2010, A7
Tags: algebra proposed, algebra
19.10.2016 17:24
Hello,I got $\boxed{\frac{1}{432}}$ wrong! Your sincerely CeuAzul
19.10.2016 19:18
I think $\frac{1}{729}$.
19.10.2016 19:26
PSY-Math wrote: I think $\frac{1}{729}$. It cant be $\frac 1{729}$ : look for example at $\left(\frac 12,\frac 1{16},...\frac 1{16}\right)$. You'll get at least one row whose product is $\frac 12\frac 1{16}\frac 1{16}$ $=\frac 1{512}>\frac 1{729}$
20.10.2016 14:17
Oh sorry I'll change my answer to $\boxed{\frac{4}{1323}}$ $\{0,\frac{1}{3},\frac{7}{27},\frac{7}{27},\frac{7}{27},\frac{7}{27},\frac{7}{27},\frac{7}{27},\frac{7}{27}\}$ Your sincerely CeuAzul
20.10.2016 14:50
CeuAzul wrote: Oh sorry I'll change my answer to $\boxed{\frac{4}{1323}}$ It cant be $\frac 1{1323}$ : look for example at $\left(\frac 12,\frac 1{16},...\frac 1{16}\right)$. You'll get at least one row whose product is $\frac 12\frac 1{16}\frac 1{16}$ $=\frac 1{512}>\frac 1{1323}$
20.10.2016 15:13
I said $\boxed{\frac{4}{1323}}$ dude Your sincerely CeuAzul
20.10.2016 15:16
CeuAzul wrote: I said $\boxed{\frac{4}{1323}}$ dude Yep, sorry
20.10.2016 16:32
Proof: WLOG $a_1 \le a_2 \le \cdots \le a_9,\sum_{i=1}^{9}a_i=1$ We form these numbers like below \[\begin{pmatrix} a_7&a_4&a_3\\a_8&a_2&a_5\\a_1&a_9&a_6 \end{pmatrix}\]Then we prove that each row & column is okay
Your sincerely CeuAzul