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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.

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

Prove the following version instead.

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.

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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.

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