Tricky AM-GM inequalityProof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$How to use induction on this type of inequality?How to prove $a_1^m + a_2^m + cdots + a_n^m geq frac1a_1 + frac1a_2 + cdots + frac1a_n$Bernoulli's inequality variationInequality $( sum_i=1^n a_i)( sum_i=1^n frac1a_i)geq n^2$Proof of this inequalityRearrangement Inequality Problem: Mathematical Olympiad.General inequalityInequality : $ (a_1a_2+a_2a_3+ldots+a_na_1)left(fraca_1a^2_2+a_2+fraca_2a^2_3+a_3+ ldots+fraca_na^2_1+a_1right)geq fracnn+1 $Regarding AM-GM inequalityInequality with two sums

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Tricky AM-GM inequality


Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$How to use induction on this type of inequality?How to prove $a_1^m + a_2^m + cdots + a_n^m geq frac1a_1 + frac1a_2 + cdots + frac1a_n$Bernoulli's inequality variationInequality $( sum_i=1^n a_i)( sum_i=1^n frac1a_i)geq n^2$Proof of this inequalityRearrangement Inequality Problem: Mathematical Olympiad.General inequalityInequality : $ (a_1a_2+a_2a_3+ldots+a_na_1)left(fraca_1a^2_2+a_2+fraca_2a^2_3+a_3+ ldots+fraca_na^2_1+a_1right)geq fracnn+1 $Regarding AM-GM inequalityInequality with two sums













3












$begingroup$


I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.



Where each $a_i$'s are positive and $n$ is a natural number.



First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if n=k+1.



Suppose the inequality holds true when n=k, i.e.,



$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.



This is true if and only if



$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.



And I got stuck here.



The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Conditions on $a_i$?
    $endgroup$
    – Parcly Taxel
    4 hours ago










  • $begingroup$
    whoa, I forgot the most important info there. They are all positive, and n is a natural number.
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 4




    $begingroup$
    Try Cauchy-Schwarz inequality?
    $endgroup$
    – Sik Feng Cheong
    4 hours ago










  • $begingroup$
    Now I get it, I learn something new every day!! Thanks a lot :D
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 2




    $begingroup$
    Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
    $endgroup$
    – Arnaud D.
    3 hours ago















3












$begingroup$


I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.



Where each $a_i$'s are positive and $n$ is a natural number.



First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if n=k+1.



Suppose the inequality holds true when n=k, i.e.,



$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.



This is true if and only if



$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.



And I got stuck here.



The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Conditions on $a_i$?
    $endgroup$
    – Parcly Taxel
    4 hours ago










  • $begingroup$
    whoa, I forgot the most important info there. They are all positive, and n is a natural number.
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 4




    $begingroup$
    Try Cauchy-Schwarz inequality?
    $endgroup$
    – Sik Feng Cheong
    4 hours ago










  • $begingroup$
    Now I get it, I learn something new every day!! Thanks a lot :D
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 2




    $begingroup$
    Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
    $endgroup$
    – Arnaud D.
    3 hours ago













3












3








3


1



$begingroup$


I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.



Where each $a_i$'s are positive and $n$ is a natural number.



First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if n=k+1.



Suppose the inequality holds true when n=k, i.e.,



$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.



This is true if and only if



$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.



And I got stuck here.



The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.










share|cite|improve this question











$endgroup$




I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.



Where each $a_i$'s are positive and $n$ is a natural number.



First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if n=k+1.



Suppose the inequality holds true when n=k, i.e.,



$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.



This is true if and only if



$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.



And I got stuck here.



The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.







analysis inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago







Ko Byeongmin

















asked 4 hours ago









Ko ByeongminKo Byeongmin

1546




1546











  • $begingroup$
    Conditions on $a_i$?
    $endgroup$
    – Parcly Taxel
    4 hours ago










  • $begingroup$
    whoa, I forgot the most important info there. They are all positive, and n is a natural number.
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 4




    $begingroup$
    Try Cauchy-Schwarz inequality?
    $endgroup$
    – Sik Feng Cheong
    4 hours ago










  • $begingroup$
    Now I get it, I learn something new every day!! Thanks a lot :D
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 2




    $begingroup$
    Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
    $endgroup$
    – Arnaud D.
    3 hours ago
















  • $begingroup$
    Conditions on $a_i$?
    $endgroup$
    – Parcly Taxel
    4 hours ago










  • $begingroup$
    whoa, I forgot the most important info there. They are all positive, and n is a natural number.
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 4




    $begingroup$
    Try Cauchy-Schwarz inequality?
    $endgroup$
    – Sik Feng Cheong
    4 hours ago










  • $begingroup$
    Now I get it, I learn something new every day!! Thanks a lot :D
    $endgroup$
    – Ko Byeongmin
    4 hours ago






  • 2




    $begingroup$
    Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
    $endgroup$
    – Arnaud D.
    3 hours ago















$begingroup$
Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago




$begingroup$
Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago












$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago




$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago




4




4




$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago




$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago












$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago




$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago




2




2




$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago




$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago










2 Answers
2






active

oldest

votes


















11












$begingroup$

Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
$$
and $$
frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
$$






share|cite|improve this answer









$endgroup$












  • $begingroup$
    That's a really strong hint, it looks like an answer
    $endgroup$
    – enedil
    2 hours ago










  • $begingroup$
    You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
    $endgroup$
    – Song
    2 hours ago


















9












$begingroup$

It is AM-HM inequality
$$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    11












    $begingroup$

    Hint: AM-GM implies
    $$
    a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
    $$
    and $$
    frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
    $$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      That's a really strong hint, it looks like an answer
      $endgroup$
      – enedil
      2 hours ago










    • $begingroup$
      You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
      $endgroup$
      – Song
      2 hours ago















    11












    $begingroup$

    Hint: AM-GM implies
    $$
    a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
    $$
    and $$
    frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
    $$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      That's a really strong hint, it looks like an answer
      $endgroup$
      – enedil
      2 hours ago










    • $begingroup$
      You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
      $endgroup$
      – Song
      2 hours ago













    11












    11








    11





    $begingroup$

    Hint: AM-GM implies
    $$
    a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
    $$
    and $$
    frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
    $$






    share|cite|improve this answer









    $endgroup$



    Hint: AM-GM implies
    $$
    a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
    $$
    and $$
    frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 4 hours ago









    SongSong

    17.2k21246




    17.2k21246











    • $begingroup$
      That's a really strong hint, it looks like an answer
      $endgroup$
      – enedil
      2 hours ago










    • $begingroup$
      You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
      $endgroup$
      – Song
      2 hours ago
















    • $begingroup$
      That's a really strong hint, it looks like an answer
      $endgroup$
      – enedil
      2 hours ago










    • $begingroup$
      You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
      $endgroup$
      – Song
      2 hours ago















    $begingroup$
    That's a really strong hint, it looks like an answer
    $endgroup$
    – enedil
    2 hours ago




    $begingroup$
    That's a really strong hint, it looks like an answer
    $endgroup$
    – enedil
    2 hours ago












    $begingroup$
    You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
    $endgroup$
    – Song
    2 hours ago




    $begingroup$
    You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
    $endgroup$
    – Song
    2 hours ago











    9












    $begingroup$

    It is AM-HM inequality
    $$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$






    share|cite|improve this answer









    $endgroup$

















      9












      $begingroup$

      It is AM-HM inequality
      $$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$






      share|cite|improve this answer









      $endgroup$















        9












        9








        9





        $begingroup$

        It is AM-HM inequality
        $$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$






        share|cite|improve this answer









        $endgroup$



        It is AM-HM inequality
        $$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        77.7k42866




        77.7k42866



























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