Why AM-GM inequality showing different results?Show that $frac a_1^2a_2+frac a_2^2a_3+…+frac a_n^2a_1geq a_1+a_2+…+a_n$ using AM-GM.If $a_1a_2cdots a_n=1$, then the sum $sum_k a_kprod_jle k (1+a_j)^-1$ is bounded below by $1-2^-n$Proof of this inequalityInequality of arithmetic means of $a_1,a_2,dots,a_n$ and $a_1,2a_2,dots,na_n$Prove that $a_1+fraca_2^2a_1+a_2+fraca_3^2a_1+a_2+a_3>b_1+fracb_2^2b_1+b_2+fracb_3^2b_1+b_2+b_3.$$a_1 + a_2 + dots a_n = 1$ find min of $a_1^2 +fraca_2^22 + dots + fraca_n^2n.$Rearrangement Inequality Problem: Mathematical Olympiad.Connection between $e^x$ convexity and inequalityMinimum value of a expression involving positive reals given their productProving that the arithmetic and geometric means of a collection of non-negative numbers lies between their minimum and maximum values

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Why AM-GM inequality showing different results?


Show that $frac a_1^2a_2+frac a_2^2a_3+…+frac a_n^2a_1geq a_1+a_2+…+a_n$ using AM-GM.If $a_1a_2cdots a_n=1$, then the sum $sum_k a_kprod_jle k (1+a_j)^-1$ is bounded below by $1-2^-n$Proof of this inequalityInequality of arithmetic means of $a_1,a_2,dots,a_n$ and $a_1,2a_2,dots,na_n$Prove that $a_1+fraca_2^2a_1+a_2+fraca_3^2a_1+a_2+a_3>b_1+fracb_2^2b_1+b_2+fracb_3^2b_1+b_2+b_3.$$a_1 + a_2 + dots a_n = 1$ find min of $a_1^2 +fraca_2^22 + dots + fraca_n^2n.$Rearrangement Inequality Problem: Mathematical Olympiad.Connection between $e^x$ convexity and inequalityMinimum value of a expression involving positive reals given their productProving that the arithmetic and geometric means of a collection of non-negative numbers lies between their minimum and maximum values













1












$begingroup$


I was given to find the minimum of $$1+a_1+a_2+a_3+...a_n$$. It was given that $$a_1timesa_2timesa_3...a_n =c$$
My approach



Using AM-GM inequality
$1+a_1+a_2+...a_nge (n+1)c^frac1n+1$




But the story doesn't ends here. There could be one more approach.

$a_1+a_2+a_3+...a_nge nc^frac1n $
So, $1+a_1+a_2+...a_nge 1+nc^frac1 n $




Both the approaches are giving different results. Is there something that I am missing?






inequality a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Note where equality is allowed in AM-GM. Is that compatible with your problem?
    $endgroup$
    – user10354138
    10 hours ago










  • $begingroup$
    In the first case equality is allowed when all the terms in the numerator equals 1. In the second case it is allowed when all the terms are equal but not necessarily 1. So which of these is correct.?
    $endgroup$
    – Abhinav
    10 hours ago
















1












$begingroup$


I was given to find the minimum of $$1+a_1+a_2+a_3+...a_n$$. It was given that $$a_1timesa_2timesa_3...a_n =c$$
My approach



Using AM-GM inequality
$1+a_1+a_2+...a_nge (n+1)c^frac1n+1$




But the story doesn't ends here. There could be one more approach.

$a_1+a_2+a_3+...a_nge nc^frac1n $
So, $1+a_1+a_2+...a_nge 1+nc^frac1 n $




Both the approaches are giving different results. Is there something that I am missing?






inequality a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Note where equality is allowed in AM-GM. Is that compatible with your problem?
    $endgroup$
    – user10354138
    10 hours ago










  • $begingroup$
    In the first case equality is allowed when all the terms in the numerator equals 1. In the second case it is allowed when all the terms are equal but not necessarily 1. So which of these is correct.?
    $endgroup$
    – Abhinav
    10 hours ago














1












1








1


1



$begingroup$


I was given to find the minimum of $$1+a_1+a_2+a_3+...a_n$$. It was given that $$a_1timesa_2timesa_3...a_n =c$$
My approach



Using AM-GM inequality
$1+a_1+a_2+...a_nge (n+1)c^frac1n+1$




But the story doesn't ends here. There could be one more approach.

$a_1+a_2+a_3+...a_nge nc^frac1n $
So, $1+a_1+a_2+...a_nge 1+nc^frac1 n $




Both the approaches are giving different results. Is there something that I am missing?






inequality a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$




I was given to find the minimum of $$1+a_1+a_2+a_3+...a_n$$. It was given that $$a_1timesa_2timesa_3...a_n =c$$
My approach



Using AM-GM inequality
$1+a_1+a_2+...a_nge (n+1)c^frac1n+1$




But the story doesn't ends here. There could be one more approach.

$a_1+a_2+a_3+...a_nge nc^frac1n $
So, $1+a_1+a_2+...a_nge 1+nc^frac1 n $




Both the approaches are giving different results. Is there something that I am missing?






inequality a.m.-g.m.-inequality




inequality a.m.-g.m.-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









HerrWarum

139117




139117










asked 10 hours ago









AbhinavAbhinav

17810




17810







  • 1




    $begingroup$
    Note where equality is allowed in AM-GM. Is that compatible with your problem?
    $endgroup$
    – user10354138
    10 hours ago










  • $begingroup$
    In the first case equality is allowed when all the terms in the numerator equals 1. In the second case it is allowed when all the terms are equal but not necessarily 1. So which of these is correct.?
    $endgroup$
    – Abhinav
    10 hours ago













  • 1




    $begingroup$
    Note where equality is allowed in AM-GM. Is that compatible with your problem?
    $endgroup$
    – user10354138
    10 hours ago










  • $begingroup$
    In the first case equality is allowed when all the terms in the numerator equals 1. In the second case it is allowed when all the terms are equal but not necessarily 1. So which of these is correct.?
    $endgroup$
    – Abhinav
    10 hours ago








1




1




$begingroup$
Note where equality is allowed in AM-GM. Is that compatible with your problem?
$endgroup$
– user10354138
10 hours ago




$begingroup$
Note where equality is allowed in AM-GM. Is that compatible with your problem?
$endgroup$
– user10354138
10 hours ago












$begingroup$
In the first case equality is allowed when all the terms in the numerator equals 1. In the second case it is allowed when all the terms are equal but not necessarily 1. So which of these is correct.?
$endgroup$
– Abhinav
10 hours ago





$begingroup$
In the first case equality is allowed when all the terms in the numerator equals 1. In the second case it is allowed when all the terms are equal but not necessarily 1. So which of these is correct.?
$endgroup$
– Abhinav
10 hours ago











2 Answers
2






active

oldest

votes


















5












$begingroup$

Both results are correct, but one of them tells you more than the other. For instance, if $x=20$, then both $xge10$ and $xge 5$ are correct, but $xge 10$ tells you more.



In your case, the second approach tells you more, because it uses explicitly the knowledge that the first term is equal to $1$. In fact, with a bit of work, you could even use this to prove that $1+nc^frac1nge(n+1)c^frac1n+1$.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Both results are correct, but they have different conditions for when the equality holds.



    For the first inequality, the equality holds when $1 = a_1 = cdots = a_n$.
    But for the second inequality, the equality holds when $a_1=cdots = a_n$.



    As TonyK mentions above, the second inequality tells you more. The equality condition would constrain all the $a_i$s and causes $c = 1 (=a_1dots a_n)$. But the second condition for the equality does not constrain any values for $c$.






    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









      5












      $begingroup$

      Both results are correct, but one of them tells you more than the other. For instance, if $x=20$, then both $xge10$ and $xge 5$ are correct, but $xge 10$ tells you more.



      In your case, the second approach tells you more, because it uses explicitly the knowledge that the first term is equal to $1$. In fact, with a bit of work, you could even use this to prove that $1+nc^frac1nge(n+1)c^frac1n+1$.






      share|cite|improve this answer









      $endgroup$

















        5












        $begingroup$

        Both results are correct, but one of them tells you more than the other. For instance, if $x=20$, then both $xge10$ and $xge 5$ are correct, but $xge 10$ tells you more.



        In your case, the second approach tells you more, because it uses explicitly the knowledge that the first term is equal to $1$. In fact, with a bit of work, you could even use this to prove that $1+nc^frac1nge(n+1)c^frac1n+1$.






        share|cite|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          Both results are correct, but one of them tells you more than the other. For instance, if $x=20$, then both $xge10$ and $xge 5$ are correct, but $xge 10$ tells you more.



          In your case, the second approach tells you more, because it uses explicitly the knowledge that the first term is equal to $1$. In fact, with a bit of work, you could even use this to prove that $1+nc^frac1nge(n+1)c^frac1n+1$.






          share|cite|improve this answer









          $endgroup$



          Both results are correct, but one of them tells you more than the other. For instance, if $x=20$, then both $xge10$ and $xge 5$ are correct, but $xge 10$ tells you more.



          In your case, the second approach tells you more, because it uses explicitly the knowledge that the first term is equal to $1$. In fact, with a bit of work, you could even use this to prove that $1+nc^frac1nge(n+1)c^frac1n+1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 10 hours ago









          TonyKTonyK

          45.2k358137




          45.2k358137





















              2












              $begingroup$

              Both results are correct, but they have different conditions for when the equality holds.



              For the first inequality, the equality holds when $1 = a_1 = cdots = a_n$.
              But for the second inequality, the equality holds when $a_1=cdots = a_n$.



              As TonyK mentions above, the second inequality tells you more. The equality condition would constrain all the $a_i$s and causes $c = 1 (=a_1dots a_n)$. But the second condition for the equality does not constrain any values for $c$.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Both results are correct, but they have different conditions for when the equality holds.



                For the first inequality, the equality holds when $1 = a_1 = cdots = a_n$.
                But for the second inequality, the equality holds when $a_1=cdots = a_n$.



                As TonyK mentions above, the second inequality tells you more. The equality condition would constrain all the $a_i$s and causes $c = 1 (=a_1dots a_n)$. But the second condition for the equality does not constrain any values for $c$.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Both results are correct, but they have different conditions for when the equality holds.



                  For the first inequality, the equality holds when $1 = a_1 = cdots = a_n$.
                  But for the second inequality, the equality holds when $a_1=cdots = a_n$.



                  As TonyK mentions above, the second inequality tells you more. The equality condition would constrain all the $a_i$s and causes $c = 1 (=a_1dots a_n)$. But the second condition for the equality does not constrain any values for $c$.






                  share|cite|improve this answer









                  $endgroup$



                  Both results are correct, but they have different conditions for when the equality holds.



                  For the first inequality, the equality holds when $1 = a_1 = cdots = a_n$.
                  But for the second inequality, the equality holds when $a_1=cdots = a_n$.



                  As TonyK mentions above, the second inequality tells you more. The equality condition would constrain all the $a_i$s and causes $c = 1 (=a_1dots a_n)$. But the second condition for the equality does not constrain any values for $c$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 9 hours ago









                  zxcvberzxcvber

                  1,368316




                  1,368316



























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