Proving Inequalities Involving Summations and Sq. Roots












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How can I prove that $sum_{i=1}^{n} |a_i| leq sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}}$ considering that $ a_{1}, a_{2}, a_{3}, ... , a_{n} $ are real numbers?



This exercise was presented in a section that also covered Cauchy Schwarz: $ (sum_{i=1}^{n}a_{i}b_{i})^2 leq (sum_{i=1}^{n}a_{i}^2)(sum_{i=1}^{n}b_{i}^2) $ but I am unsure if anything related to Cauchy Schwarz is involved in this proof.



On the RHS, I proceeded with $ sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}} $ = $ sqrt{nsum_{i=1}^{n} a_{i}^{2} } $ but then I'm stuck and unsure how to proceed. Any hints/help in this direction is greatly appreciated.



Thank you!






real-analysis inequality proof-writing cauchy-schwarz-inequality







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  • 4




    $begingroup$
    Write $|a_i| = 1cdot |a_i|$ and apply C-S.
    $endgroup$
    – Song
    Jan 11 at 21:44
















0












$begingroup$


How can I prove that $sum_{i=1}^{n} |a_i| leq sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}}$ considering that $ a_{1}, a_{2}, a_{3}, ... , a_{n} $ are real numbers?



This exercise was presented in a section that also covered Cauchy Schwarz: $ (sum_{i=1}^{n}a_{i}b_{i})^2 leq (sum_{i=1}^{n}a_{i}^2)(sum_{i=1}^{n}b_{i}^2) $ but I am unsure if anything related to Cauchy Schwarz is involved in this proof.



On the RHS, I proceeded with $ sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}} $ = $ sqrt{nsum_{i=1}^{n} a_{i}^{2} } $ but then I'm stuck and unsure how to proceed. Any hints/help in this direction is greatly appreciated.



Thank you!






real-analysis inequality proof-writing cauchy-schwarz-inequality







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    Write $|a_i| = 1cdot |a_i|$ and apply C-S.
    $endgroup$
    – Song
    Jan 11 at 21:44














0












0








0





$begingroup$


How can I prove that $sum_{i=1}^{n} |a_i| leq sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}}$ considering that $ a_{1}, a_{2}, a_{3}, ... , a_{n} $ are real numbers?



This exercise was presented in a section that also covered Cauchy Schwarz: $ (sum_{i=1}^{n}a_{i}b_{i})^2 leq (sum_{i=1}^{n}a_{i}^2)(sum_{i=1}^{n}b_{i}^2) $ but I am unsure if anything related to Cauchy Schwarz is involved in this proof.



On the RHS, I proceeded with $ sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}} $ = $ sqrt{nsum_{i=1}^{n} a_{i}^{2} } $ but then I'm stuck and unsure how to proceed. Any hints/help in this direction is greatly appreciated.



Thank you!






real-analysis inequality proof-writing cauchy-schwarz-inequality







share|cite|improve this question









$endgroup$




How can I prove that $sum_{i=1}^{n} |a_i| leq sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}}$ considering that $ a_{1}, a_{2}, a_{3}, ... , a_{n} $ are real numbers?



This exercise was presented in a section that also covered Cauchy Schwarz: $ (sum_{i=1}^{n}a_{i}b_{i})^2 leq (sum_{i=1}^{n}a_{i}^2)(sum_{i=1}^{n}b_{i}^2) $ but I am unsure if anything related to Cauchy Schwarz is involved in this proof.



On the RHS, I proceeded with $ sqrt{n} sqrt{sum_{i=1}^{n} a_{i}^{2}} $ = $ sqrt{nsum_{i=1}^{n} a_{i}^{2} } $ but then I'm stuck and unsure how to proceed. Any hints/help in this direction is greatly appreciated.



Thank you!






real-analysis inequality proof-writing cauchy-schwarz-inequality




real-analysis inequality proof-writing cauchy-schwarz-inequality






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asked Jan 11 at 21:42









Zen'zZen'z

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243








  • 4




    $begingroup$
    Write $|a_i| = 1cdot |a_i|$ and apply C-S.
    $endgroup$
    – Song
    Jan 11 at 21:44














  • 4




    $begingroup$
    Write $|a_i| = 1cdot |a_i|$ and apply C-S.
    $endgroup$
    – Song
    Jan 11 at 21:44








4




4




$begingroup$
Write $|a_i| = 1cdot |a_i|$ and apply C-S.
$endgroup$
– Song
Jan 11 at 21:44




$begingroup$
Write $|a_i| = 1cdot |a_i|$ and apply C-S.
$endgroup$
– Song
Jan 11 at 21:44










2 Answers
2






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1












$begingroup$

Let $b_i =1$ for $1le ile n$



Then apply Cauchy Schwartz inequality to $|a_i|$ and your $b_i=1$



You get the result right away upon taking square root of both sides.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    It's
    $$sqrt{(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)}geq$$
    $$geqsqrt{1^2cdot a_1^2}+sqrt{1^2cdot a_2^2}+...sqrt{1^2cdot a_n^2}=|a_1|+|a_2|+...+|a_n|.$$






    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









      1












      $begingroup$

      Let $b_i =1$ for $1le ile n$



      Then apply Cauchy Schwartz inequality to $|a_i|$ and your $b_i=1$



      You get the result right away upon taking square root of both sides.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $b_i =1$ for $1le ile n$



        Then apply Cauchy Schwartz inequality to $|a_i|$ and your $b_i=1$



        You get the result right away upon taking square root of both sides.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $b_i =1$ for $1le ile n$



          Then apply Cauchy Schwartz inequality to $|a_i|$ and your $b_i=1$



          You get the result right away upon taking square root of both sides.






          share|cite|improve this answer









          $endgroup$



          Let $b_i =1$ for $1le ile n$



          Then apply Cauchy Schwartz inequality to $|a_i|$ and your $b_i=1$



          You get the result right away upon taking square root of both sides.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 22:16









          Mohammad Riazi-KermaniMohammad Riazi-Kermani

          41.6k42061




          41.6k42061























              1












              $begingroup$

              It's
              $$sqrt{(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)}geq$$
              $$geqsqrt{1^2cdot a_1^2}+sqrt{1^2cdot a_2^2}+...sqrt{1^2cdot a_n^2}=|a_1|+|a_2|+...+|a_n|.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                It's
                $$sqrt{(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)}geq$$
                $$geqsqrt{1^2cdot a_1^2}+sqrt{1^2cdot a_2^2}+...sqrt{1^2cdot a_n^2}=|a_1|+|a_2|+...+|a_n|.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  It's
                  $$sqrt{(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)}geq$$
                  $$geqsqrt{1^2cdot a_1^2}+sqrt{1^2cdot a_2^2}+...sqrt{1^2cdot a_n^2}=|a_1|+|a_2|+...+|a_n|.$$






                  share|cite|improve this answer









                  $endgroup$



                  It's
                  $$sqrt{(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)}geq$$
                  $$geqsqrt{1^2cdot a_1^2}+sqrt{1^2cdot a_2^2}+...sqrt{1^2cdot a_n^2}=|a_1|+|a_2|+...+|a_n|.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 11 at 22:17









                  Michael RozenbergMichael Rozenberg

                  103k1791195




                  103k1791195






























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