Extending Cauchy-Schwarz to any $p in (1,infty)$












0












$begingroup$


Given $n$ non-negative real number $leftlbrace a_irightrbrace_{i=1}^n$, we know that using Cauchy-Schwarz, one can prove that
$$
sumlimits_{i=1}^n sqrt{a_i} leq sqrt{n} cdot sqrt{sumlimits_{i=1}^n a_i}
$$



Is it true that the same would apply to any $p in (1,infty)$, i.e.
$$
sumlimits_{i=1}^n sqrt[p]{a_i} leq sqrt[p]{n} cdot sqrt[p]{sumlimits_{i=1}^n a_i} ?
$$



If not, can we multiply the right term by some constant depending on $p$ which will result in having the same form of the inequality?



Please advise.






calculus linear-algebra inequality roots cauchy-schwarz-inequality







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  • $begingroup$
    See en.wikipedia.org/wiki/…
    $endgroup$
    – GEdgar
    Jan 30 at 13:12










  • $begingroup$
    @ClementC. I might be missing something, but isn't he applying Cauchy-Schwarz to $sqrt{a_i} cdot 1$, resulting in $sqrt{a_i}^2 = a_i$?
    $endgroup$
    – Klaus
    Jan 30 at 13:17












  • $begingroup$
    @Klaus My bad, early morning... you're right.
    $endgroup$
    – Clement C.
    Jan 30 at 13:18
















0












$begingroup$


Given $n$ non-negative real number $leftlbrace a_irightrbrace_{i=1}^n$, we know that using Cauchy-Schwarz, one can prove that
$$
sumlimits_{i=1}^n sqrt{a_i} leq sqrt{n} cdot sqrt{sumlimits_{i=1}^n a_i}
$$



Is it true that the same would apply to any $p in (1,infty)$, i.e.
$$
sumlimits_{i=1}^n sqrt[p]{a_i} leq sqrt[p]{n} cdot sqrt[p]{sumlimits_{i=1}^n a_i} ?
$$



If not, can we multiply the right term by some constant depending on $p$ which will result in having the same form of the inequality?



Please advise.






calculus linear-algebra inequality roots cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    See en.wikipedia.org/wiki/…
    $endgroup$
    – GEdgar
    Jan 30 at 13:12










  • $begingroup$
    @ClementC. I might be missing something, but isn't he applying Cauchy-Schwarz to $sqrt{a_i} cdot 1$, resulting in $sqrt{a_i}^2 = a_i$?
    $endgroup$
    – Klaus
    Jan 30 at 13:17












  • $begingroup$
    @Klaus My bad, early morning... you're right.
    $endgroup$
    – Clement C.
    Jan 30 at 13:18














0












0








0





$begingroup$


Given $n$ non-negative real number $leftlbrace a_irightrbrace_{i=1}^n$, we know that using Cauchy-Schwarz, one can prove that
$$
sumlimits_{i=1}^n sqrt{a_i} leq sqrt{n} cdot sqrt{sumlimits_{i=1}^n a_i}
$$



Is it true that the same would apply to any $p in (1,infty)$, i.e.
$$
sumlimits_{i=1}^n sqrt[p]{a_i} leq sqrt[p]{n} cdot sqrt[p]{sumlimits_{i=1}^n a_i} ?
$$



If not, can we multiply the right term by some constant depending on $p$ which will result in having the same form of the inequality?



Please advise.






calculus linear-algebra inequality roots cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$




Given $n$ non-negative real number $leftlbrace a_irightrbrace_{i=1}^n$, we know that using Cauchy-Schwarz, one can prove that
$$
sumlimits_{i=1}^n sqrt{a_i} leq sqrt{n} cdot sqrt{sumlimits_{i=1}^n a_i}
$$



Is it true that the same would apply to any $p in (1,infty)$, i.e.
$$
sumlimits_{i=1}^n sqrt[p]{a_i} leq sqrt[p]{n} cdot sqrt[p]{sumlimits_{i=1}^n a_i} ?
$$



If not, can we multiply the right term by some constant depending on $p$ which will result in having the same form of the inequality?



Please advise.






calculus linear-algebra inequality roots cauchy-schwarz-inequality




calculus linear-algebra inequality roots cauchy-schwarz-inequality






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share|cite|improve this question













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edited Jan 30 at 13:21









Bernard

124k741117




124k741117










asked Jan 30 at 13:10









user3492773user3492773

19915




19915












  • $begingroup$
    See en.wikipedia.org/wiki/…
    $endgroup$
    – GEdgar
    Jan 30 at 13:12










  • $begingroup$
    @ClementC. I might be missing something, but isn't he applying Cauchy-Schwarz to $sqrt{a_i} cdot 1$, resulting in $sqrt{a_i}^2 = a_i$?
    $endgroup$
    – Klaus
    Jan 30 at 13:17












  • $begingroup$
    @Klaus My bad, early morning... you're right.
    $endgroup$
    – Clement C.
    Jan 30 at 13:18


















  • $begingroup$
    See en.wikipedia.org/wiki/…
    $endgroup$
    – GEdgar
    Jan 30 at 13:12










  • $begingroup$
    @ClementC. I might be missing something, but isn't he applying Cauchy-Schwarz to $sqrt{a_i} cdot 1$, resulting in $sqrt{a_i}^2 = a_i$?
    $endgroup$
    – Klaus
    Jan 30 at 13:17












  • $begingroup$
    @Klaus My bad, early morning... you're right.
    $endgroup$
    – Clement C.
    Jan 30 at 13:18
















$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– GEdgar
Jan 30 at 13:12




$begingroup$
See en.wikipedia.org/wiki/…
$endgroup$
– GEdgar
Jan 30 at 13:12












$begingroup$
@ClementC. I might be missing something, but isn't he applying Cauchy-Schwarz to $sqrt{a_i} cdot 1$, resulting in $sqrt{a_i}^2 = a_i$?
$endgroup$
– Klaus
Jan 30 at 13:17






$begingroup$
@ClementC. I might be missing something, but isn't he applying Cauchy-Schwarz to $sqrt{a_i} cdot 1$, resulting in $sqrt{a_i}^2 = a_i$?
$endgroup$
– Klaus
Jan 30 at 13:17














$begingroup$
@Klaus My bad, early morning... you're right.
$endgroup$
– Clement C.
Jan 30 at 13:18




$begingroup$
@Klaus My bad, early morning... you're right.
$endgroup$
– Clement C.
Jan 30 at 13:18










3 Answers
3






active

oldest

votes


















2












$begingroup$

You are probably looking for https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am familiar with Hölder's inequality, yet how is this connected to here? I need a bound on the sum of $p$-roots using the $p$-root of the sum.
    $endgroup$
    – user3492773
    Jan 30 at 13:21






  • 1




    $begingroup$
    Just do the same thing as you did with Cauchy-Schwarz. In fact, Cauchy-Schwarz is a special case of Hölder.
    $endgroup$
    – Klaus
    Jan 30 at 13:22





















2












$begingroup$

By Holder's inequality, for $frac{1}{p}+frac{1}{q}=1$, $1le p,q$, $$sum_{i=1}^nsqrt[p]{a_i}le|1|_q|(sqrt[p]{a_i})|_p=sqrt[q]{n}sqrt[p]{sum_{i=1}^na_i}$$ Note that the root power of $n$ is not $p$ (except when $p=2$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I just proved it using Klaus's suggestion, yet i thank you!!!
    $endgroup$
    – user3492773
    Jan 30 at 13:35



















2












$begingroup$

It's wrong.



Try $n=p=3$ and $a_1=a_2=a_3=1.$



We need to prove that $$3leqsqrt[3]3cdotsqrt[3]3,$$ which is impossible.



By the way, by Holder
$$n^{p-1}sum_{k=1}^na_kgeqleft(sum_{k=1}^nleft(a_kright)^{frac{1}{p}}right)^p,$$ which gives $$n^{1-frac{1}{p}}left(sum_{k=1}^na_kright)^{frac{1}{p}}geqsum_{k=1}^nleft(a_kright)^{frac{1}{p}}
.$$






share|cite|improve this answer











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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You are probably looking for https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am familiar with Hölder's inequality, yet how is this connected to here? I need a bound on the sum of $p$-roots using the $p$-root of the sum.
      $endgroup$
      – user3492773
      Jan 30 at 13:21






    • 1




      $begingroup$
      Just do the same thing as you did with Cauchy-Schwarz. In fact, Cauchy-Schwarz is a special case of Hölder.
      $endgroup$
      – Klaus
      Jan 30 at 13:22


















    2












    $begingroup$

    You are probably looking for https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am familiar with Hölder's inequality, yet how is this connected to here? I need a bound on the sum of $p$-roots using the $p$-root of the sum.
      $endgroup$
      – user3492773
      Jan 30 at 13:21






    • 1




      $begingroup$
      Just do the same thing as you did with Cauchy-Schwarz. In fact, Cauchy-Schwarz is a special case of Hölder.
      $endgroup$
      – Klaus
      Jan 30 at 13:22
















    2












    2








    2





    $begingroup$

    You are probably looking for https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure.






    share|cite|improve this answer









    $endgroup$



    You are probably looking for https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 at 13:11









    KlausKlaus

    2,907214




    2,907214












    • $begingroup$
      I am familiar with Hölder's inequality, yet how is this connected to here? I need a bound on the sum of $p$-roots using the $p$-root of the sum.
      $endgroup$
      – user3492773
      Jan 30 at 13:21






    • 1




      $begingroup$
      Just do the same thing as you did with Cauchy-Schwarz. In fact, Cauchy-Schwarz is a special case of Hölder.
      $endgroup$
      – Klaus
      Jan 30 at 13:22




















    • $begingroup$
      I am familiar with Hölder's inequality, yet how is this connected to here? I need a bound on the sum of $p$-roots using the $p$-root of the sum.
      $endgroup$
      – user3492773
      Jan 30 at 13:21






    • 1




      $begingroup$
      Just do the same thing as you did with Cauchy-Schwarz. In fact, Cauchy-Schwarz is a special case of Hölder.
      $endgroup$
      – Klaus
      Jan 30 at 13:22


















    $begingroup$
    I am familiar with Hölder's inequality, yet how is this connected to here? I need a bound on the sum of $p$-roots using the $p$-root of the sum.
    $endgroup$
    – user3492773
    Jan 30 at 13:21




    $begingroup$
    I am familiar with Hölder's inequality, yet how is this connected to here? I need a bound on the sum of $p$-roots using the $p$-root of the sum.
    $endgroup$
    – user3492773
    Jan 30 at 13:21




    1




    1




    $begingroup$
    Just do the same thing as you did with Cauchy-Schwarz. In fact, Cauchy-Schwarz is a special case of Hölder.
    $endgroup$
    – Klaus
    Jan 30 at 13:22






    $begingroup$
    Just do the same thing as you did with Cauchy-Schwarz. In fact, Cauchy-Schwarz is a special case of Hölder.
    $endgroup$
    – Klaus
    Jan 30 at 13:22













    2












    $begingroup$

    By Holder's inequality, for $frac{1}{p}+frac{1}{q}=1$, $1le p,q$, $$sum_{i=1}^nsqrt[p]{a_i}le|1|_q|(sqrt[p]{a_i})|_p=sqrt[q]{n}sqrt[p]{sum_{i=1}^na_i}$$ Note that the root power of $n$ is not $p$ (except when $p=2$).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I just proved it using Klaus's suggestion, yet i thank you!!!
      $endgroup$
      – user3492773
      Jan 30 at 13:35
















    2












    $begingroup$

    By Holder's inequality, for $frac{1}{p}+frac{1}{q}=1$, $1le p,q$, $$sum_{i=1}^nsqrt[p]{a_i}le|1|_q|(sqrt[p]{a_i})|_p=sqrt[q]{n}sqrt[p]{sum_{i=1}^na_i}$$ Note that the root power of $n$ is not $p$ (except when $p=2$).






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I just proved it using Klaus's suggestion, yet i thank you!!!
      $endgroup$
      – user3492773
      Jan 30 at 13:35














    2












    2








    2





    $begingroup$

    By Holder's inequality, for $frac{1}{p}+frac{1}{q}=1$, $1le p,q$, $$sum_{i=1}^nsqrt[p]{a_i}le|1|_q|(sqrt[p]{a_i})|_p=sqrt[q]{n}sqrt[p]{sum_{i=1}^na_i}$$ Note that the root power of $n$ is not $p$ (except when $p=2$).






    share|cite|improve this answer









    $endgroup$



    By Holder's inequality, for $frac{1}{p}+frac{1}{q}=1$, $1le p,q$, $$sum_{i=1}^nsqrt[p]{a_i}le|1|_q|(sqrt[p]{a_i})|_p=sqrt[q]{n}sqrt[p]{sum_{i=1}^na_i}$$ Note that the root power of $n$ is not $p$ (except when $p=2$).







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 at 13:30









    ChrystomathChrystomath

    1,978513




    1,978513












    • $begingroup$
      I just proved it using Klaus's suggestion, yet i thank you!!!
      $endgroup$
      – user3492773
      Jan 30 at 13:35


















    • $begingroup$
      I just proved it using Klaus's suggestion, yet i thank you!!!
      $endgroup$
      – user3492773
      Jan 30 at 13:35
















    $begingroup$
    I just proved it using Klaus's suggestion, yet i thank you!!!
    $endgroup$
    – user3492773
    Jan 30 at 13:35




    $begingroup$
    I just proved it using Klaus's suggestion, yet i thank you!!!
    $endgroup$
    – user3492773
    Jan 30 at 13:35











    2












    $begingroup$

    It's wrong.



    Try $n=p=3$ and $a_1=a_2=a_3=1.$



    We need to prove that $$3leqsqrt[3]3cdotsqrt[3]3,$$ which is impossible.



    By the way, by Holder
    $$n^{p-1}sum_{k=1}^na_kgeqleft(sum_{k=1}^nleft(a_kright)^{frac{1}{p}}right)^p,$$ which gives $$n^{1-frac{1}{p}}left(sum_{k=1}^na_kright)^{frac{1}{p}}geqsum_{k=1}^nleft(a_kright)^{frac{1}{p}}
    .$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      It's wrong.



      Try $n=p=3$ and $a_1=a_2=a_3=1.$



      We need to prove that $$3leqsqrt[3]3cdotsqrt[3]3,$$ which is impossible.



      By the way, by Holder
      $$n^{p-1}sum_{k=1}^na_kgeqleft(sum_{k=1}^nleft(a_kright)^{frac{1}{p}}right)^p,$$ which gives $$n^{1-frac{1}{p}}left(sum_{k=1}^na_kright)^{frac{1}{p}}geqsum_{k=1}^nleft(a_kright)^{frac{1}{p}}
      .$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        It's wrong.



        Try $n=p=3$ and $a_1=a_2=a_3=1.$



        We need to prove that $$3leqsqrt[3]3cdotsqrt[3]3,$$ which is impossible.



        By the way, by Holder
        $$n^{p-1}sum_{k=1}^na_kgeqleft(sum_{k=1}^nleft(a_kright)^{frac{1}{p}}right)^p,$$ which gives $$n^{1-frac{1}{p}}left(sum_{k=1}^na_kright)^{frac{1}{p}}geqsum_{k=1}^nleft(a_kright)^{frac{1}{p}}
        .$$






        share|cite|improve this answer











        $endgroup$



        It's wrong.



        Try $n=p=3$ and $a_1=a_2=a_3=1.$



        We need to prove that $$3leqsqrt[3]3cdotsqrt[3]3,$$ which is impossible.



        By the way, by Holder
        $$n^{p-1}sum_{k=1}^na_kgeqleft(sum_{k=1}^nleft(a_kright)^{frac{1}{p}}right)^p,$$ which gives $$n^{1-frac{1}{p}}left(sum_{k=1}^na_kright)^{frac{1}{p}}geqsum_{k=1}^nleft(a_kright)^{frac{1}{p}}
        .$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 30 at 15:48

























        answered Jan 30 at 15:26









        Michael RozenbergMichael Rozenberg

        109k1896201




        109k1896201






























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