Proof $mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$












0












$begingroup$


We have $(a + b)^{n} geq a^{n} + b^{n}$



Proof:



$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$






binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $a$ and $b$ ? What is $n$ ?
    $endgroup$
    – Rebellos
    Dec 9 '18 at 19:48










  • $begingroup$
    a&b positive real numbers, n natural number*
    $endgroup$
    – Lamine
    Dec 9 '18 at 19:57
















0












$begingroup$


We have $(a + b)^{n} geq a^{n} + b^{n}$



Proof:



$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$






binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $a$ and $b$ ? What is $n$ ?
    $endgroup$
    – Rebellos
    Dec 9 '18 at 19:48










  • $begingroup$
    a&b positive real numbers, n natural number*
    $endgroup$
    – Lamine
    Dec 9 '18 at 19:57














0












0








0





$begingroup$


We have $(a + b)^{n} geq a^{n} + b^{n}$



Proof:



$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$






binomial-theorem







share|cite|improve this question











$endgroup$




We have $(a + b)^{n} geq a^{n} + b^{n}$



Proof:



$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$






binomial-theorem




binomial-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 19:36







Lamine

















asked Dec 9 '18 at 19:27









LamineLamine

12




12












  • $begingroup$
    What is $a$ and $b$ ? What is $n$ ?
    $endgroup$
    – Rebellos
    Dec 9 '18 at 19:48










  • $begingroup$
    a&b positive real numbers, n natural number*
    $endgroup$
    – Lamine
    Dec 9 '18 at 19:57


















  • $begingroup$
    What is $a$ and $b$ ? What is $n$ ?
    $endgroup$
    – Rebellos
    Dec 9 '18 at 19:48










  • $begingroup$
    a&b positive real numbers, n natural number*
    $endgroup$
    – Lamine
    Dec 9 '18 at 19:57
















$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48




$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48












$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57




$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57










2 Answers
2






active

oldest

votes


















0












$begingroup$

You can show this by using induction.



https://en.wikipedia.org/wiki/Mathematical_induction#Description



I'll leave it to you to verify it using a base.



Now assume that



$mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)



is true. We want to deduce that



$mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)



is true as well.



Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Raising to the $n$-th power,
    $|a - b|^{frac{1}{n}}
    geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
    $
    becomes
    $|a - b|
    geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
    $    .



    Letting
    $a = c^n$ and $b = d^n$,
    this becomes
    $|c^n - d^n|
    geq || c| - |d||^n
    $    .



    Assuming that
    $|c| ge |d|$,
    dividing by $c^n$
    and letting
    $frac{d}{c} = x$,
    this is
    $|1 - x^n|
    geq |1 - x|^n
    $    .



    Assume that $1 ge x ge 0$.
    So we want to show that
    $1-x^n ge (1-x)^n$.



    This is true for $x=0$
    and $x=1$.



    This is true for $n=1$.



    Since $x^n$ and $(1-x)^n$
    are decreasing
    with increasing $n$,
    $1-x^n$ is increasing and
    $(1-x)^n$ is decreasing
    so the inequality holds
    for all $n$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      You can show this by using induction.



      https://en.wikipedia.org/wiki/Mathematical_induction#Description



      I'll leave it to you to verify it using a base.



      Now assume that



      $mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)



      is true. We want to deduce that



      $mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)



      is true as well.



      Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        You can show this by using induction.



        https://en.wikipedia.org/wiki/Mathematical_induction#Description



        I'll leave it to you to verify it using a base.



        Now assume that



        $mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)



        is true. We want to deduce that



        $mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)



        is true as well.



        Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          You can show this by using induction.



          https://en.wikipedia.org/wiki/Mathematical_induction#Description



          I'll leave it to you to verify it using a base.



          Now assume that



          $mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)



          is true. We want to deduce that



          $mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)



          is true as well.



          Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?






          share|cite|improve this answer









          $endgroup$



          You can show this by using induction.



          https://en.wikipedia.org/wiki/Mathematical_induction#Description



          I'll leave it to you to verify it using a base.



          Now assume that



          $mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)



          is true. We want to deduce that



          $mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)



          is true as well.



          Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 19:43









          CruZCruZ

          638




          638























              0












              $begingroup$

              Raising to the $n$-th power,
              $|a - b|^{frac{1}{n}}
              geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
              $
              becomes
              $|a - b|
              geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
              $              .



              Letting
              $a = c^n$ and $b = d^n$,
              this becomes
              $|c^n - d^n|
              geq || c| - |d||^n
              $              .



              Assuming that
              $|c| ge |d|$,
              dividing by $c^n$
              and letting
              $frac{d}{c} = x$,
              this is
              $|1 - x^n|
              geq |1 - x|^n
              $              .



              Assume that $1 ge x ge 0$.
              So we want to show that
              $1-x^n ge (1-x)^n$.



              This is true for $x=0$
              and $x=1$.



              This is true for $n=1$.



              Since $x^n$ and $(1-x)^n$
              are decreasing
              with increasing $n$,
              $1-x^n$ is increasing and
              $(1-x)^n$ is decreasing
              so the inequality holds
              for all $n$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Raising to the $n$-th power,
                $|a - b|^{frac{1}{n}}
                geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
                $
                becomes
                $|a - b|
                geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
                $                .



                Letting
                $a = c^n$ and $b = d^n$,
                this becomes
                $|c^n - d^n|
                geq || c| - |d||^n
                $                .



                Assuming that
                $|c| ge |d|$,
                dividing by $c^n$
                and letting
                $frac{d}{c} = x$,
                this is
                $|1 - x^n|
                geq |1 - x|^n
                $                .



                Assume that $1 ge x ge 0$.
                So we want to show that
                $1-x^n ge (1-x)^n$.



                This is true for $x=0$
                and $x=1$.



                This is true for $n=1$.



                Since $x^n$ and $(1-x)^n$
                are decreasing
                with increasing $n$,
                $1-x^n$ is increasing and
                $(1-x)^n$ is decreasing
                so the inequality holds
                for all $n$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Raising to the $n$-th power,
                  $|a - b|^{frac{1}{n}}
                  geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
                  $
                  becomes
                  $|a - b|
                  geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
                  $                  .



                  Letting
                  $a = c^n$ and $b = d^n$,
                  this becomes
                  $|c^n - d^n|
                  geq || c| - |d||^n
                  $                  .



                  Assuming that
                  $|c| ge |d|$,
                  dividing by $c^n$
                  and letting
                  $frac{d}{c} = x$,
                  this is
                  $|1 - x^n|
                  geq |1 - x|^n
                  $                  .



                  Assume that $1 ge x ge 0$.
                  So we want to show that
                  $1-x^n ge (1-x)^n$.



                  This is true for $x=0$
                  and $x=1$.



                  This is true for $n=1$.



                  Since $x^n$ and $(1-x)^n$
                  are decreasing
                  with increasing $n$,
                  $1-x^n$ is increasing and
                  $(1-x)^n$ is decreasing
                  so the inequality holds
                  for all $n$.






                  share|cite|improve this answer









                  $endgroup$



                  Raising to the $n$-th power,
                  $|a - b|^{frac{1}{n}}
                  geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
                  $
                  becomes
                  $|a - b|
                  geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
                  $                  .



                  Letting
                  $a = c^n$ and $b = d^n$,
                  this becomes
                  $|c^n - d^n|
                  geq || c| - |d||^n
                  $                  .



                  Assuming that
                  $|c| ge |d|$,
                  dividing by $c^n$
                  and letting
                  $frac{d}{c} = x$,
                  this is
                  $|1 - x^n|
                  geq |1 - x|^n
                  $                  .



                  Assume that $1 ge x ge 0$.
                  So we want to show that
                  $1-x^n ge (1-x)^n$.



                  This is true for $x=0$
                  and $x=1$.



                  This is true for $n=1$.



                  Since $x^n$ and $(1-x)^n$
                  are decreasing
                  with increasing $n$,
                  $1-x^n$ is increasing and
                  $(1-x)^n$ is decreasing
                  so the inequality holds
                  for all $n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 20:49









                  marty cohenmarty cohen

                  74.5k549129




                  74.5k549129






























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