Prove that $x_1x_2x_3\cdots x_n>y_1y_2y_3\cdots y_m$ with some given conditions

Asked May 31 '14 at 03:53

Active May 31 '14 at 06:42

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$x_1, x_2, x_3,.\cdots ,x_n$ and $y_1, y_2, y_3,\ldots,y_m$ are two series of positive integers, such that $m\not=n$.

Given that $1<x_1< x_2<x_3<\cdots<x_n<y_1< y_2< y_3<\cdots<y_m$ and $x_1+ x_2+ x_3+\cdots+x_n > y_1+ y_2+ y_3+\cdots+y_m$, prove that $x_1x_2x_3\cdots x_n>y_1y_2y_3\cdots y_m$.

I have tried to prove it for conservatives. I have no idea how to deal with this type of inequality properly.

edited May 31 '14 at 05:43

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asked May 31 '14 at 03:53

Satvik Mashkaria

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Is there a reason you think it is true? What do you mean "conservatives?" – Thomas Andrews May 31 '14 at 04:13
Did you try induction? – user21820 May 31 '14 at 04:22
did you see this problem from a math journal? – DeepSea May 31 '14 at 04:28
Do you have $m<n$ ? Otherwise there is no solution.
If $1<x_1<x_2<y_1<y_2<y_3$, then $(x_1+x_2)<(y_1+y_1)<(y_1+y_2+y_3)$
– Orca May 31 '14 at 04:44
I think it's implied that $m<n$ since $x_1+\cdots+ x_n>y_1+\cdots +y_m$. – Rocket Man May 31 '14 at 04:52
@Orca $m<n$ is guaranteed by the additive condition (along with the strictly increasing condition). – Peter Woolfitt May 31 '14 at 04:52
I think so: $nx_n\ge x_1+\cdots+x_n>y_1+\cdots+y_m\ge my_1$, and $x_n<y_1$. – peterwhy May 31 '14 at 04:53
I think this is http://en.wikipedia.org/wiki/Muirhead%27s_inequality – math110 May 31 '14 at 05:08
@user21820 I don't know even how to solve this with the help of induction. – Satvik Mashkaria May 31 '14 at 07:57
Not just that $m<n$, $m<\frac{\sum_{i=1}^n x_1}{x_n}$.
In fact $m$ is even more severely restricted.

Does this help ?
– Orca May 31 '14 at 08:18
Sorry the solution I thought I had in my head vanished with a puff. – user21820 May 31 '14 at 08:37

Prove that $x_1x_2x_3\cdots x_n>y_1y_2y_3\cdots y_m$ with some given conditions

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