Let $n\ge 2$ be an integer. $a_1,a_2,\dotsc,a_n$ are arbitrarily chosen positive integers with $(a_1,a_2,\dotsc,a_n)=1$. Let $A=a_1+a_2+\dotsb+a_n$ and $(A,a_i)=d_i$. Let $(a_2,a_3,\dotsc,a_n)=D_1, (a_1,a_3,\dotsc,a_n)=D_2,\dotsc, (a_1,a_2,\dotsc,a_{n-1})=D_n$. Find the minimum of $\prod\limits_{i=1}^n\dfrac{A-a_i}{d_iD_i}$
Problem
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Tags: number theory proposed, number theory
apluscactus
05.04.2013 16:50
It is at least 1, because each member of the product is at least one. But can someone propose a construction to show 1 works ?
phoenixAspies
05.04.2013 18:25
There has been a solution, on sqing's sina blog.
s372102
06.04.2013 19:44
phoenixAspies wrote: There has been a solution, on sqing's sina blog. Did you mean http://blog.sina.com.cn/s/blog_4c1131020101701e.html ? I don't see a solution
vntbqpqh234
22.04.2013 11:19
I think the answer is $(n-1)^{n}$ Have that: $(D_{i},D_{j})=1$ all $i<>j$ then we have $a_{j}=r_{j}.D_{1}.D_{2}...D_{n}/ D_{j}$ and $(r_{j},D_{i})=1$ all $i <> j$ Hence $d_{i}=(A-a_{i},a_{i})=(A-a_{i},r_{i})$ then $d_{i}\leq r_{i}$ And use AM-GM, $\frac{A-a_{i}}{d_{i}D_{i}} \geq (n-1).\frac{ (r_{1}.r_{2}..r_{n}/r_{i})^{1/(n-1)}.(D_{1}.D_{2}...D_{n}/D_{i})^{(n-2)/(n-1)}}{r_{i}}$(*) With (*) we have min is $(n-1)^{n}$ and have that when all $a_{i}=1$
vntbqpqh234
22.04.2013 11:20
I think the answer is $(n-1)^{n}$ Have that: $(D_{i},D_{j})=1$ all $i<>j$ then we have $a_{j}=r_{j}.D_{1}.D_{2}...D_{n}/ D_{j}$ and $r_{j},D_{i}=1$ all $i<>j$ Hence $d_{i}=(A-a_{i},a_{i})=(A-a_{i},r_{i})$ then $d_{i}\leq r_{i}$ And use AM-GM, $\frac{A-a_{i}}{d_{i}D_{i}} \geq (n-1).\frac{ (r_{1}.r_{2}..r_{n}/r_{i})^{1/(n-1)}.(D_{1}.D_{2}...D_{n}/D_{i})^{(n-2)/(n-1)}}{r_{i}}$(*) With (*) we have min is $(n-1)^{n}$ and have that when all $a_{i}=1$
JG666
09.04.2022 02:35
Nice problem combined ineq with NT~
We claim the minimum value is $(n-1)^n.$
Let $d=\gcd(d_i,D_{i+1}).$ Then, $d\mid a_1,a_2,\cdots,a_i,a_{i+2},\cdots,a_n,d\mid a_1+a_2+\cdots+a_n.$
yielding $d\mid a_1,a_2,\cdots,a_i,\cdots,a_n\Rightarrow d=1.$
We conclude that for any $i\ne j\in \mathbb N_+, \gcd(d_i,D_j)=1.$
Thus, $d_{i-1}D_i\mid a_{i-1}\Rightarrow d_{i-1}D_i\leqslant a_{i-1}\Rightarrow \prod_{i=1}^nd_iD_i\leqslant \prod_{i=1}^na_i.$
By AM-GM, $\prod_{i=1}^n(A-a_i)\geqslant (n-1)^n\prod_{i=1}^na_i.\quad\square$
ChanandlerBong
05.07.2023 17:19
Solution: We claim the answer is $(n-1)^n$. Notice that $\text{gcd}(D_i,D_j)=\text{gcd}(a_1,\dots,a_n)=1$ for $1\leq i<j\leq n$. Because $a_k$ is divided by all $D_j$ where $j\neq k$, then $\prod\limits_{j\neq k}D_j \mid a_k$. Set $a_k:=x_k\prod\limits_{j\neq k}D_j$. It is obvious that $\text{gcd}(A,D_i)=1$, therefore $d_i\mid x_i$ for $1\leq i \leq n$. Thus $$\prod\limits_{i=1}^n\dfrac{A-a_i}{d_iD_i}\geq\prod\limits_{i=1}^{n}\frac{\sum\limits_{j\neq i}x_j}{x_i}\geq(n-1)^n$$the last inequality is due to AM-GM inequality. Also $a_1=\dots=a_n=1$ is an equality case.