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









$endgroup$












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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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            2017 IndyCar Series Contents Series news Teams and drivers Schedule Season summary Footnotes References External links Navigation menu"INDYCAR: Initial 2018 bodywork concepts unveiled"the original"IndyCar confirms switch to Performance Friction brakes in 2017""AJ Foyt Racing will switch to Chevy"the original"Carlos Munoz, Conor Daly will drive for AJ Foyt Racing""Zach Veach's Indy 500 Debut Confirmed with Foyt""No mass exodus from Honda after Ganassi switch""Ex-F1 driver Sato joins Andretti Autosport for 2017 IndyCar season""IndyCar's Ryan Hunter-Reay, sponsor DHL paired through 2020""hhgregg and Andretti Autosport announce partnership for key races in 2016""INDYCAR: Rossi re-signs with Andretti"the original"McLaren Formula 1 - Fernando Alonso to race at Indy 500 with McLaren, Honda and Andretti Autosport""Shank will finally take part in Indy 500 with Harvey, Andretti | MotorSportsTalk""Andretti adds Jack Harvey to Indy 500 field""Ganassi switches to Honda power for 2017""INDYCAR: Chilton returns to Ganassi"the original"IndyCar silly season: Who's going where in 2017?""INDYCAR: Kanaan, NTT Data return to Ganassi"the original"Kimball to remain at Ganassi for 2017""Coyne confirms Bourdais for 2017 IndyCar season""Davison to sub for Bourdais in Indy 500"the original"Gutierrez confirmed for Detroit IndyCar debut""Gutierrez returns with Coyne for rest of 2017 season""Vautier to drive for Coyne at Texas"the original"INDYCAR: Coyne confirms Jones for 2017"the original"Pippa Mann returns to Coyne for Indy 500""Karam, Dreyer & Reinbold teaming up again for Indianapolis 500""Pigot to return to Ed Carpenter Racing""Hildebrand confirmed as full-time Ed Carpenter driver""Veach to replace injured Hildebrand at Barber"the originalNew Team Harding Racing Enters Chaves for 101st Indianapolis 500"Juncos Racing Announces Entry in 101st Running of the Indianapolis 500 :: Juncos Racing""Juncos confirms Pigot for Indy 500""Saavedra confirmed in Juncos' second 500 entry"the original"Lazier confirms Indy 500 run after son's USF2000 debut"the original"Claman DeMelo to race for RLLR at Sonoma"the original"Rahal signs Servia and ace engineer for 2017""IndyCar: Aleshin returns with Schmidt"the original"Aleshin replaced by Saavedra for Toronto""Jack Harvey will pilot SPM No. 7 car at Watkins Glen, Sonoma""Jay Howard confirmed in Tony Stewart's supported SPM Indy entry""INDYCAR: Newgarden to wave the flag at Penske"the original"Pagenaud opts for No. 1 in 2017"the original"Penske confirms Newgarden for 2017""Montoya to stay with Team Penske in 2017""Target leaving IndyCar after 27 seasons with Chip Ganassi""Cavin: IndyCar could see complete driver/team shakeup in 2017""End of the road for KV Racing?""KV Racing confirms closure, equipment sold to Juncos""Juncos confirms IndyCar Series entry"the original"Juncos readies IndyCar program, aims for '17 500"the original"Harding Racing to add Texas, Pocono to schedule"the original"Sato signs with Andretti Autosport for 2017""INDYCAR: Aleshin in Doubt at SPM"the original"Long Beach notebook: JR Hildebrand breaks hand""Hildebrand cleared to return at Phoenix"the original"Bourdais to undergo surgery on multiple fractures""Aleshin loses Schmidt Peterson IndyCar ride""Saavedra in at SPM for Pocono, Gateway"the original"Bourdais to make return at Gateway"the original"The IndyCar Grand Prix no longer is sponsored by Angie's List""2017 IndyCar Series rulebook""2017 Verizon IndyCar Series Official Rulebook"Official websiteeeeee