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Math Help - Prove n^2<= ......

  1. #1
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    Prove n^2<= ......

    Hi,

    Much appreciate if anyone could help with this...

    Prove

    n^2 <= (a1+...+an).(1/a1+...+1/an)
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  2. #2
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    hi
    (a_{1}+....+a_{n})(\frac{1}{a_{1}}+...+\frac{1}{a_  {n}})=\sum_{k=1}^{n}a_{k}\sum_{k=1}^{n}\frac{1}{a_  {k}}=\sum_{k=1}^{n}1=n
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  3. #3
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    Quote Originally Posted by Raoh View Post
    hi
    (a_{1}+....+a_{n})(\frac{1}{a_{1}}+...+\frac{1}{a_  {n}})=\sum_{k=1}^{n}a_{k}\sum_{k=1}^{n}\frac{1}{a_  {k}}=\sum_{k=1}^{n}1=n
    OH!
    (1+2)\left(1+\frac{1}{2}\right)=\frac{9}{2}.
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  4. #4
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    Quote Originally Posted by Plato View Post
    OH!
    (1+2)\left(1+\frac{1}{2}\right)=\frac{9}{2}.
    hhh ! where i went wrong ?
    thanks (i think i'm dizzy )
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  5. #5
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    Quote Originally Posted by chings View Post
    Hi,

    Much appreciate if anyone could help with this...

    Prove

    n^2 <= (a1+...+an).(1/a1+...+1/an)

    It must be a_i>0\,\,\,\forall\,\,i, otherwise 2^2\nleq (1+(-1))\left(\frac{1}{1}+\frac{-1}{1}\right)=0

    Now, do you know the Means Inequalities?:

    \frac{a_1+...+a_n}{n}\,\,\geq \,\,\sqrt[n]{a_1\cdot ...\cdot a_n}\,\,\geq\,\, \frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}}

    Well, taking the two extremes in the above inequalities gives you what you want. About the proof of the inequalities (well, THE inequality, since only the left one is needed: the right one follows from this taking the inverse of the elements) you can search inside MHF and look for a solution. I sent one just yesterday, in another thread)

    Tonio
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  6. #6
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    Quote Originally Posted by Raoh View Post
    hhh ! where i went wrong ?
    thanks (i think i'm dizzy )
    Well, what you said is just not right :x

    \sum_k a_k \cdot \sum_i b_i = a_1(\sum_i b_i) + a_2(\sum_i b_i) + ... \neq a_1b_1 + a_2b_2 + ...
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  7. #7
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    Quote Originally Posted by Defunkt View Post
    Well, what you said is just not right :x

    \sum_k a_k \cdot \sum_i b_i = a_1(\sum_i b_i) + a_2(\sum_i b_i) + ... \neq a_1b_1 + a_2b_2 + ...
    thank you.
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  8. #8
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    Quote Originally Posted by tonio View Post
    It must be a_i>0\,\,\,\forall\,\,i, otherwise 2^2\nleq (1+(-1))\left(\frac{1}{1}+\frac{-1}{1}\right)=0

    Now, do you know the Means Inequalities?:

    \frac{a_1+...+a_n}{n}\,\,\geq \,\,\sqrt[n]{a_1\cdot ...\cdot a_n}\,\,\geq\,\, \frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}}

    Well, taking the two extremes in the above inequalities gives you what you want. About the proof of the inequalities (well, THE inequality, since only the left one is needed: the right one follows from this taking the inverse of the elements) you can search inside MHF and look for a solution. I sent one just yesterday, in another thread)

    Tonio

    Hi all,

    Thanks for the responses. Tonio: don't know much about the mean inequalities but yes indeed, ai are positive real numbers. and begging your pardon but what is MHF? I'll search for the post for the proof for the above.
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  9. #9
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    Quote Originally Posted by chings View Post
    ... pardon but what is MHF? .
    I'll give you a hint, it's big, blue, and you just logged onto it.
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  10. #10
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    zomg noobed.

    thanks again.
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  11. #11
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    Hi Tonio,

    still can't seem to find the page. Could you kindly assist please thank you.
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  12. #12
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    Quote Originally Posted by chings View Post
    Hi Tonio,

    still can't seem to find the page. Could you kindly assist please thank you.
    you mean MHF ?
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  13. #13
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    Quote Originally Posted by chings View Post
    Hi Tonio,

    still can't seem to find the page. Could you kindly assist please thank you.

    It's here: http://www.mathhelpforum.com/math-he...-exercise.html

    The second post there is mine and where I prove the inequality.

    Tonio
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