anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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linda3
  • linda3
Whats your question?
anonymous
  • anonymous
If \[a _{i}>0 \forall i \in N\] such that \[\prod_{n}^{i=1}a _{i}=1\] Then prove that \[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}).....(1+a _{n}) \ge 2^{n}\]
anonymous
  • anonymous
@ikram002p @ganeshie8

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anonymous
  • anonymous
What does the sign inverted u mean
ganeshie8
  • ganeshie8
product of terms
ikram002p
  • ikram002p
ive seen this before :O
ganeshie8
  • ganeshie8
its like \(\sum \), \[\large \prod_{k=1}^{3}k= (1)(2)(3)\]
ganeshie8
  • ganeshie8
\[\large \prod_{k=1}^{3}k^2= (1^2)(2^2)(3^2)\]
ganeshie8
  • ganeshie8
etc..
anonymous
  • anonymous
oh
ganeshie8
  • ganeshie8
the proof for the original question is simple
anonymous
  • anonymous
Yes then it is easy after i understood what the sign means
anonymous
  • anonymous
i will be back after dinner
ganeshie8
  • ganeshie8
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
  • ikram002p
mmm sure its a proof ?
ikram002p
  • ikram002p
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
  • ikram002p
so \(0
ikram002p
  • ikram002p
wait it can be rational but not integer
ikram002p
  • ikram002p
forget about the interval
ganeshie8
  • ganeshie8
\[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}) \cdots (1+a _{n}) = \prod \limits_{i=1}^{n} (a_i + 1)\]
ganeshie8
  • ganeshie8
right ?
ikram002p
  • ikram002p
yep
ganeshie8
  • ganeshie8
ahh i see your point, so we're given that \(a_i \gt 0\), we're NOT given that its a natural number
ikram002p
  • ikram002p
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
  • ganeshie8
okay then it cannot be an one liner >.<
ikram002p
  • ikram002p
\(a_i \) 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
  • ganeshie8
lets expand the product
ikram002p
  • ikram002p
the thing is im too hungry lol xD so quick i wanna sleep hehe unless there is a given condition for \(a_i\)
anonymous
  • anonymous
I got it how to do
ikram002p
  • ikram002p
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
  • ikram002p
only this case work , if \(1\le a_i \) (1+3/2) (1+5/4) (1+7/6) >=2^3
ikram002p
  • ikram002p
nope we cant , we only can conclude that \(a_i\) is rational
ikram002p
  • ikram002p
for example \(\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{1}×\dfrac{1}{1}=1\)
ganeshie8
  • ganeshie8
ok
ikram002p
  • ikram002p
gtg nw , have a nice day
anonymous
  • anonymous
|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
  • ganeshie8
nice :) AM/GM inequality to the rescue !!
anonymous
  • anonymous
yes, just didn't knew what that sign meant
ikram002p
  • ikram002p
cool O.O never seen this before
ganeshie8
  • ganeshie8
AM-GM inequality : \(\dfrac{a+b}{2} \ge \sqrt{ab}\)
ganeshie8
  • ganeshie8
@ikram002p
ikram002p
  • ikram002p
ic .. i dont think that ive seen it before :D

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