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No.name

  • 5 months ago

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  1. linda3
    • 5 months ago
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    Whats your question?

  2. No.name
    • 5 months ago
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    If \[a _{i}>0 \forall i \in N\] such that \[\prod_{n}^{i=1}a _{i}=1\] Then prove that \[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}).....(1+a _{n}) \ge 2^{n}\]

  3. No.name
    • 5 months ago
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    @ikram002p @ganeshie8

  4. No.name
    • 5 months ago
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    What does the sign inverted u mean

  5. ganeshie8
    • 5 months ago
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    product of terms

  6. ikram002p
    • 5 months ago
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    ive seen this before :O

  7. ganeshie8
    • 5 months ago
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    its like \(\sum \), \[\large \prod_{k=1}^{3}k= (1)(2)(3)\]

  8. ganeshie8
    • 5 months ago
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    \[\large \prod_{k=1}^{3}k^2= (1^2)(2^2)(3^2)\]

  9. ganeshie8
    • 5 months ago
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    etc..

  10. No.name
    • 5 months ago
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    oh

  11. ganeshie8
    • 5 months ago
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    the proof for the original question is simple

  12. No.name
    • 5 months ago
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    Yes then it is easy after i understood what the sign means

  13. No.name
    • 5 months ago
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    i will be back after dinner

  14. ganeshie8
    • 5 months ago
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    yes :) here is an one line proof : since \(a_i \in \mathbb{N}\), \(a_i+1 \ge 2 \implies \prod \limits_{i=1}^{n}(a _{i}+1)\ge 2^n \)

  15. ikram002p
    • 5 months ago
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    mmm sure its a proof ?

  16. ikram002p
    • 5 months ago
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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

  17. ikram002p
    • 5 months ago
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    so \(0<a_i\le 1\) do u agree to this @ganeshie8 ?

  18. ikram002p
    • 5 months ago
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    wait it can be rational but not integer

  19. ikram002p
    • 5 months ago
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    forget about the interval

  20. ganeshie8
    • 5 months ago
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    \[(1+a _{1})(1+a _{2})(1+a _{3})(1+a _{4}) \cdots (1+a _{n}) = \prod \limits_{i=1}^{n} (a_i + 1)\]

  21. ganeshie8
    • 5 months ago
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    right ?

  22. ikram002p
    • 5 months ago
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    yep

  23. ganeshie8
    • 5 months ago
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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

  24. ikram002p
    • 5 months ago
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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

  25. ganeshie8
    • 5 months ago
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    okay then it cannot be an one liner >.<

  26. ikram002p
    • 5 months ago
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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

  27. ganeshie8
    • 5 months ago
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    lets expand the product

  28. ikram002p
    • 5 months ago
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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 \(a_i\)

  29. No.name
    • 5 months ago
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    I got it how to do

  30. ikram002p
    • 5 months ago
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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)

  31. ikram002p
    • 5 months ago
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    only this case work , if \(1\le a_i \) (1+3/2) (1+5/4) (1+7/6) >=2^3

  32. ikram002p
    • 5 months ago
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    nope we cant , we only can conclude that \(a_i\) is rational

  33. ikram002p
    • 5 months ago
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    for example \(\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{3}{1}×\dfrac{1}{1}=1\)

  34. ganeshie8
    • 5 months ago
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    ok

  35. ikram002p
    • 5 months ago
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    gtg nw , have a nice day

  36. No.name
    • 5 months ago
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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

  37. ganeshie8
    • 5 months ago
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    nice :) AM/GM inequality to the rescue !!

  38. No.name
    • 5 months ago
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    yes, just didn't knew what that sign meant

  39. ikram002p
    • 5 months ago
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    cool O.O never seen this before

  40. ganeshie8
    • 5 months ago
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    AM-GM inequality : \(\dfrac{a+b}{2} \ge \sqrt{ab}\)

  41. ganeshie8
    • 5 months ago
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    @ikram002p

  42. ikram002p
    • 5 months ago
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    ic .. i dont think that ive seen it before :D

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