Prove that: $x_1\cdot x_2\cdots x_n>y_1\cdot y_2\cdots y_m$.

For two positive integer sequences $x_1,x_2,\ldots,x_n$ and $y_1,y_2,\ldots,y_m$ satisfying

• $x_i\neq x_j\quad \text{and}\quad y_i\neq y_j\quad \forall i,j, i \ne j$

• $1<x_1<x_2<\cdots<x_n<y_1<\cdots<y_m.$

• $x_1+x_2+\cdots+x_n>y_1+\cdots+y_m.$

Prove that: $x_1\cdot x_2\cdots x_n>y_1\cdot y_2\cdots y_m$.

(from internet)

I don't have an idea for this problem. Thanks for your help.

• facebook.com/… – user41499 May 11 '14 at 23:22
• this question come from internet? Do you not have any references? – leticia May 12 '14 at 13:20
• I find this problem on facebook. – user41499 May 12 '14 at 14:38
• @leticia see here; it means "Original Poster". – 6005 May 12 '14 at 20:41
• I think one should use Induction. – Math137 May 16 '14 at 16:30

Here's my approach.

For two positive integer $x_1,x_2,...,x_n$ and $y_1,y_2,...,y_m$ which satisfy

1. $x_i \neq x_j\quad \text{and}\quad y_i\neq y_j\quad \forall 1 < i < j$,

2. $1<x_2<...<x_n<y_2<...<y_m,$ and $1 \leq x_1$ and $1 \leq y_1$

3. $x_1+x_2+ \ldots +x_n > y_1+ \ldots +y_m.$

Prove that: $x_1 \times x_2 \times \cdots \times x_n \geq y_1 \times y_2 \times \cdots y_m$.

This version is much easier to work with. We then prove strict inequality by looking at the equality cases.

Hint: Think about what $x_1, y_1$ could be made to do.
Hint: How would you minimize the LHS and maximize the RHS?
Hint: Deal with $m=1$ separately, which results in the equality case. The original version for $m=1$ is straightforward.

Rename it into one sequence:

Write $x_i=1+\sum_{k=1}^{i} a_k$, that is $x_1=1+a_1$,$x_2=x_1+a_2$ etc.

and $y_i=1+\sum_{k=1+n}^{i+n} a_k$

Your second hypothesis implies the first one, and it says $$\forall\, i \quad a_i>0.$$

Your third hypothesis is $$\sum_{i=1}^n(1+\sum_{k=1}^i a_i) > \sum_{i=1+n}^{m+n}(1+\sum_{k=1}^i a_i),$$ in other words

$$n+na_1+(n-1)a_2+\ldots+a_n>m+m(a_1+\ldots+a_n)+(m-1)a_{n+1}+\ldots+a_{n+m}$$ and you want to find the minimum of $$\Pi_{i=1}^n\left(1+\sum_{k=1}^i a_i\right)-\Pi_{i=1+n}^{n+m}\left(1+\sum_{k=1}^{i+n} a_i\right)$$ Looks like a good'ole Lagrange multiplier problem.