prove that $displaystyle n^nbigg(frac{n+1}{2}bigg)^{2n}geq (n!)^3$ for natural number $n$












1












$begingroup$


prove that $$n^nbigg(frac{n+1}{2}bigg)^{2n}geq (n!)^3.$$ for $n$ is a natural number



what i try



i have use AM GM inequality



$$frac{1^3+2^3+3^3+cdots +n^3}{n}geq ((n!))^{frac{1}{3n}}$$



how i prove question inequality help me please






inequality







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  • $begingroup$
    Hint: $1^3+2^3+...+n^3=frac{n(n+1)(2n+1)}{6}$
    $endgroup$
    – Peter Foreman
    Feb 2 at 13:55










  • $begingroup$
    That is the sum of first natural squares I think not of cubes
    $endgroup$
    – Thomas
    Feb 2 at 16:42
















1












$begingroup$


prove that $$n^nbigg(frac{n+1}{2}bigg)^{2n}geq (n!)^3.$$ for $n$ is a natural number



what i try



i have use AM GM inequality



$$frac{1^3+2^3+3^3+cdots +n^3}{n}geq ((n!))^{frac{1}{3n}}$$



how i prove question inequality help me please






inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: $1^3+2^3+...+n^3=frac{n(n+1)(2n+1)}{6}$
    $endgroup$
    – Peter Foreman
    Feb 2 at 13:55










  • $begingroup$
    That is the sum of first natural squares I think not of cubes
    $endgroup$
    – Thomas
    Feb 2 at 16:42














1












1








1





$begingroup$


prove that $$n^nbigg(frac{n+1}{2}bigg)^{2n}geq (n!)^3.$$ for $n$ is a natural number



what i try



i have use AM GM inequality



$$frac{1^3+2^3+3^3+cdots +n^3}{n}geq ((n!))^{frac{1}{3n}}$$



how i prove question inequality help me please






inequality







share|cite|improve this question











$endgroup$




prove that $$n^nbigg(frac{n+1}{2}bigg)^{2n}geq (n!)^3.$$ for $n$ is a natural number



what i try



i have use AM GM inequality



$$frac{1^3+2^3+3^3+cdots +n^3}{n}geq ((n!))^{frac{1}{3n}}$$



how i prove question inequality help me please






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 at 14:09









Bernard

124k741117




124k741117










asked Feb 2 at 13:46









jackyjacky

1,349816




1,349816












  • $begingroup$
    Hint: $1^3+2^3+...+n^3=frac{n(n+1)(2n+1)}{6}$
    $endgroup$
    – Peter Foreman
    Feb 2 at 13:55










  • $begingroup$
    That is the sum of first natural squares I think not of cubes
    $endgroup$
    – Thomas
    Feb 2 at 16:42


















  • $begingroup$
    Hint: $1^3+2^3+...+n^3=frac{n(n+1)(2n+1)}{6}$
    $endgroup$
    – Peter Foreman
    Feb 2 at 13:55










  • $begingroup$
    That is the sum of first natural squares I think not of cubes
    $endgroup$
    – Thomas
    Feb 2 at 16:42
















$begingroup$
Hint: $1^3+2^3+...+n^3=frac{n(n+1)(2n+1)}{6}$
$endgroup$
– Peter Foreman
Feb 2 at 13:55




$begingroup$
Hint: $1^3+2^3+...+n^3=frac{n(n+1)(2n+1)}{6}$
$endgroup$
– Peter Foreman
Feb 2 at 13:55












$begingroup$
That is the sum of first natural squares I think not of cubes
$endgroup$
– Thomas
Feb 2 at 16:42




$begingroup$
That is the sum of first natural squares I think not of cubes
$endgroup$
– Thomas
Feb 2 at 16:42










2 Answers
2






active

oldest

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3












$begingroup$

First o all, by expanding the terms we obtain$$n^nbigg(frac{n+1}{2}bigg)^{2n}{={1over n^n}cdot bigg(frac{n(n+1)}{2}bigg)^{2n}\={1over n^n}cdot (1+2^3+cdots +n^3)^n}$$by substitution and taking the $n$-th root, we need to prove that $${1+2^3+cdots+n^3over n}ge sqrt[n]{n!^3}=sqrt[n]{1^3cdot 2^3 cdots n^3}$$which is straight-forward by AM-GM






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You can also use Stirling approximation and Taylor expansions to show that
    $$a_n=n log (n)+2 n log left(frac{n+1}{2}right) - 3 log(n!) >0$$ This would give
    $$a_n=left(2-frac{3}{2} log
    left({2 pi }right)right)+ (3-log (4)),n-frac{3}{2} log left({n}right)-frac{5}{4
    n}+Oleft(frac{1}{n^2}right)$$     which is positive $forall, n geq 2$.






    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









      3












      $begingroup$

      First o all, by expanding the terms we obtain$$n^nbigg(frac{n+1}{2}bigg)^{2n}{={1over n^n}cdot bigg(frac{n(n+1)}{2}bigg)^{2n}\={1over n^n}cdot (1+2^3+cdots +n^3)^n}$$by substitution and taking the $n$-th root, we need to prove that $${1+2^3+cdots+n^3over n}ge sqrt[n]{n!^3}=sqrt[n]{1^3cdot 2^3 cdots n^3}$$which is straight-forward by AM-GM






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        First o all, by expanding the terms we obtain$$n^nbigg(frac{n+1}{2}bigg)^{2n}{={1over n^n}cdot bigg(frac{n(n+1)}{2}bigg)^{2n}\={1over n^n}cdot (1+2^3+cdots +n^3)^n}$$by substitution and taking the $n$-th root, we need to prove that $${1+2^3+cdots+n^3over n}ge sqrt[n]{n!^3}=sqrt[n]{1^3cdot 2^3 cdots n^3}$$which is straight-forward by AM-GM






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          First o all, by expanding the terms we obtain$$n^nbigg(frac{n+1}{2}bigg)^{2n}{={1over n^n}cdot bigg(frac{n(n+1)}{2}bigg)^{2n}\={1over n^n}cdot (1+2^3+cdots +n^3)^n}$$by substitution and taking the $n$-th root, we need to prove that $${1+2^3+cdots+n^3over n}ge sqrt[n]{n!^3}=sqrt[n]{1^3cdot 2^3 cdots n^3}$$which is straight-forward by AM-GM






          share|cite|improve this answer









          $endgroup$



          First o all, by expanding the terms we obtain$$n^nbigg(frac{n+1}{2}bigg)^{2n}{={1over n^n}cdot bigg(frac{n(n+1)}{2}bigg)^{2n}\={1over n^n}cdot (1+2^3+cdots +n^3)^n}$$by substitution and taking the $n$-th root, we need to prove that $${1+2^3+cdots+n^3over n}ge sqrt[n]{n!^3}=sqrt[n]{1^3cdot 2^3 cdots n^3}$$which is straight-forward by AM-GM







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 2 at 14:34









          Mostafa AyazMostafa Ayaz

          18.1k31040




          18.1k31040























              1












              $begingroup$

              You can also use Stirling approximation and Taylor expansions to show that
              $$a_n=n log (n)+2 n log left(frac{n+1}{2}right) - 3 log(n!) >0$$ This would give
              $$a_n=left(2-frac{3}{2} log
              left({2 pi }right)right)+ (3-log (4)),n-frac{3}{2} log left({n}right)-frac{5}{4
              n}+Oleft(frac{1}{n^2}right)$$               which is positive $forall, n geq 2$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                You can also use Stirling approximation and Taylor expansions to show that
                $$a_n=n log (n)+2 n log left(frac{n+1}{2}right) - 3 log(n!) >0$$ This would give
                $$a_n=left(2-frac{3}{2} log
                left({2 pi }right)right)+ (3-log (4)),n-frac{3}{2} log left({n}right)-frac{5}{4
                n}+Oleft(frac{1}{n^2}right)$$                 which is positive $forall, n geq 2$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You can also use Stirling approximation and Taylor expansions to show that
                  $$a_n=n log (n)+2 n log left(frac{n+1}{2}right) - 3 log(n!) >0$$ This would give
                  $$a_n=left(2-frac{3}{2} log
                  left({2 pi }right)right)+ (3-log (4)),n-frac{3}{2} log left({n}right)-frac{5}{4
                  n}+Oleft(frac{1}{n^2}right)$$                   which is positive $forall, n geq 2$.






                  share|cite|improve this answer











                  $endgroup$



                  You can also use Stirling approximation and Taylor expansions to show that
                  $$a_n=n log (n)+2 n log left(frac{n+1}{2}right) - 3 log(n!) >0$$ This would give
                  $$a_n=left(2-frac{3}{2} log
                  left({2 pi }right)right)+ (3-log (4)),n-frac{3}{2} log left({n}right)-frac{5}{4
                  n}+Oleft(frac{1}{n^2}right)$$                   which is positive $forall, n geq 2$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 3 at 3:25

























                  answered Feb 2 at 16:13









                  Claude LeiboviciClaude Leibovici

                  125k1158135




                  125k1158135






























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