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linda3
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Whats your question?
No.name
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If \[a _{i}>0 \forall i \in N\]
such that \[\prod_{n}^{i=1}a _{i}=1\]
Then prove that \[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}).....(1+a _{n}) \ge 2^{n}\]
No.name
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@ikram002p @ganeshie8
No.name
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What does the sign inverted u mean
ganeshie8
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product of terms
ikram002p
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ive seen this before :O
ganeshie8
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its like \(\sum \),
\[\large \prod_{k=1}^{3}k= (1)(2)(3)\]
ganeshie8
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\[\large \prod_{k=1}^{3}k^2= (1^2)(2^2)(3^2)\]
ganeshie8
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etc..
No.name
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oh
ganeshie8
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the proof for the original question is simple
No.name
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Yes then it is easy after i understood what the sign means
No.name
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i will be back after dinner
ganeshie8
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yes :) here is an one line proof :
since \(a_i \in \mathbb{N}\), \(a_i+1 \ge 2 \implies \prod \limits_{i=1}^{n}(a _{i}+1)\ge 2^n \)
ikram002p
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mmm sure its a proof ?
ikram002p
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i thought i had to do this
\(\prod_{n}^{i=1}a _{i}\prod_{n}^{i=1} (1+\dfrac{1}{a_i}) \)
but i got confused for a second sense\( a_i \) is not integer check the product =1
ikram002p
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so \(0<a_i\le 1\)
do u agree to this @ganeshie8 ?
ikram002p
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wait it can be rational but not integer
ikram002p
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forget about the interval
ganeshie8
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\[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}) \cdots (1+a _{n}) = \prod \limits_{i=1}^{n} (a_i + 1)\]
ganeshie8
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right ?
ikram002p
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yep
ganeshie8
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ahh i see your point, so we're given that \(a_i \gt 0\),
we're NOT given that its a natural number
ikram002p
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yep exactly and sence product of a_i = 1
then for sure it cant be integers
but also it can be <1 or >1
ganeshie8
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okay then it cannot be an one liner >.<
ikram002p
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\(a_i \) should be rationl
even though its confusing
in that case what if \(a_i \le \) 1
then problem solved
but if \(a_i >1 \)like 3/2 or 5/2
then we can do the first line proof u have
but what if \(a_i \) was mixed btw number which >1 or <1 or =1
then we can't conclude anything in this case
ganeshie8
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lets expand the product
ikram002p
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the thing is im too hungry lol xD
so quick i wanna sleep hehe
unless there is a given condition for \(a_i\)
No.name
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I got it how to do
ikram002p
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see ganesh
for example
(1+1/2) (1+1/3) (1+1/4) <2^3
(1+1/3776568)(1+3/2)(1+1/2)(1/45456) <2^4 ( not sure :P)
ikram002p
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only this case work , if \(1\le a_i \)
(1+3/2) (1+5/4) (1+7/6) >=2^3
ikram002p
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nope we cant , we only can conclude that \(a_i\) is rational
ikram002p
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for example
\(\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{1}×\dfrac{1}{1}=1\)
ganeshie8
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ok
ikram002p
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gtg nw ,
have a nice day
No.name
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|dw:1404314484947:dw|
\[(1+a _{1})(1+a _{2})(1+a _{3}).....(1+a _{n})\ge2^{n}[(a _{1}a _{2}a _{3}..a _{n}]^{1/2}\]
|dw:1404314863576:dw|
so proved
ganeshie8
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nice :) AM/GM inequality to the rescue !!
No.name
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yes, just didn't knew what that sign meant
ikram002p
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cool O.O
never seen this before
ganeshie8
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AM-GM inequality : \(\dfrac{a+b}{2} \ge \sqrt{ab}\)
ganeshie8
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@ikram002p
ikram002p
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ic .. i dont think that ive seen it before :D