Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies...












0












$begingroup$


Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions







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  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22
















0












$begingroup$


Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22














0












0








0





$begingroup$


Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions







share|cite|improve this question











$endgroup$




Prove the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$ satisfies $a_{n} > 2$




Let $x = a_{n}/2$.



Then $a_{n + 1} = x + 1/x$.



Define $f(x) = x + 1/x$ so that $f'(x) = -1/x^2 + 1 = 0 implies x = 1,$ meaning that $a_{n + 1}$ has a minimum at $1 + 1/1 = 2$.



This shows $a_{n + 1} geq 2$. But I want to show the strict bound $a_{n + 1} > 2$. Perhaps I can prove the case separately. I can't make any progress.






real-analysis calculus sequences-and-series inequality monotone-functions




real-analysis calculus sequences-and-series inequality monotone-functions






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edited Dec 12 '18 at 6:31







stackofhay42

















asked Dec 12 '18 at 5:37









stackofhay42stackofhay42

1956




1956








  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22














  • 1




    $begingroup$
    Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
    $endgroup$
    – Lau
    Dec 12 '18 at 5:43












  • $begingroup$
    All these and a lot more are covered here, including linked questions ...
    $endgroup$
    – rtybase
    Dec 12 '18 at 9:22








1




1




$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43






$begingroup$
Prove $a_n>0$ and use $a+bgeq 2sqrt{ab}$.
$endgroup$
– Lau
Dec 12 '18 at 5:43














$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22




$begingroup$
All these and a lot more are covered here, including linked questions ...
$endgroup$
– rtybase
Dec 12 '18 at 9:22










4 Answers
4






active

oldest

votes


















4












$begingroup$

Use principle of mathematical induction,



Basis for induction
$$a_1>2$$



Induction Hypothesis
$$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



Inductive step
$$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
which is true by condition of equality in AM-GM inequality and induction hypothesis.



Hence proved



Hope it is helpful






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    AM-GM gives:
    $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



    Edit after comments:




    • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

    • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


    So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



    For a direct proof that all members of the iteration are greater than $2$ you may consider




    • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


    Now use MVT:



    $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      how do i know $a neq 2$?
      $endgroup$
      – stackofhay42
      Dec 12 '18 at 6:31










    • $begingroup$
      @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
      $endgroup$
      – Yadati Kiran
      Dec 12 '18 at 6:47










    • $begingroup$
      Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
      $endgroup$
      – stackofhay42
      Dec 12 '18 at 6:51












    • $begingroup$
      @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
      $endgroup$
      – Martund
      Dec 12 '18 at 9:25










    • $begingroup$
      @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
      $endgroup$
      – Martund
      Dec 12 '18 at 9:27





















    2












    $begingroup$

    Note $a_n >0$, $n=1,2,3,.....$.



    Let $n ge 1$:



    $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



    $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



    $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



    (The first term $ge 0$, a square divided by a positive number.)



    Hence



    $a_{n+1} ge 2$, $n in mathbb{N}$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Following with your reasoning, suppose $x + 1/x = 2$. Then:
      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
      But this is impossible because...






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        What about $a_n<2$?
        $endgroup$
        – Rakesh Bhatt
        Dec 12 '18 at 10:14










      • $begingroup$
        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
        $endgroup$
        – Martín-Blas Pérez Pinilla
        Dec 12 '18 at 10:43












      • $begingroup$
        Why is this impossible?
        $endgroup$
        – Ekesh Kumar
        Dec 13 '18 at 23:45










      • $begingroup$
        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
        $endgroup$
        – Martín-Blas Pérez Pinilla
        Dec 14 '18 at 7:12











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Use principle of mathematical induction,



      Basis for induction
      $$a_1>2$$



      Induction Hypothesis
      $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



      Inductive step
      $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
      which is true by condition of equality in AM-GM inequality and induction hypothesis.



      Hence proved



      Hope it is helpful






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Use principle of mathematical induction,



        Basis for induction
        $$a_1>2$$



        Induction Hypothesis
        $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



        Inductive step
        $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
        which is true by condition of equality in AM-GM inequality and induction hypothesis.



        Hence proved



        Hope it is helpful






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Use principle of mathematical induction,



          Basis for induction
          $$a_1>2$$



          Induction Hypothesis
          $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



          Inductive step
          $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
          which is true by condition of equality in AM-GM inequality and induction hypothesis.



          Hence proved



          Hope it is helpful






          share|cite|improve this answer









          $endgroup$



          Use principle of mathematical induction,



          Basis for induction
          $$a_1>2$$



          Induction Hypothesis
          $$a_k>2$$ $$Longleftrightarrow frac{a_k}{2}>1$$



          Inductive step
          $$a_{k+1}=frac{a_k}{2}+frac{2}{a_k}>2$$
          which is true by condition of equality in AM-GM inequality and induction hypothesis.



          Hence proved



          Hope it is helpful







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 6:08









          MartundMartund

          1,623213




          1,623213























              4












              $begingroup$

              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27


















              4












              $begingroup$

              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27
















              4












              4








              4





              $begingroup$

              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$






              share|cite|improve this answer











              $endgroup$



              AM-GM gives:
              $$frac{a}{2} + frac{2}{a} stackrel{a>0, aneq 2}{>} 2sqrt{frac{a}{2}cdotfrac{2}{a}} = 2$$



              Edit after comments:




              • Note that $a^{star} = 2$ is a fixpoint of the iteration as $frac{a^{star}}{2} + frac{2}{a^{star}} = 2$.

              • For any other starting value $a_0 > 0, a_0 neq 2$ AM-GM tells you, that the iteration $a_{n+1} = frac{a_n}{2} + frac{2}{a_n}$ produces $a_{n+1} > 2$.


              So, also in your specific case of $a_1 = 4$ all members of the iteration are greater than $2$.



              For a direct proof that all members of the iteration are greater than $2$ you may consider




              • $f(x) = frac{x}{2} + frac{2}{x} Rightarrow f'(x) = frac{1}{2} - frac{2}{x^2}$


              Now use MVT:



              $$f(x) - 2 = f(x) - f(2) = f'(xi)(x - 2) = left(frac{1}{2} - frac{2}{{xi}^2}right) (x - 2) stackrel{2 < xi < x}{>} 0$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 12 '18 at 9:35

























              answered Dec 12 '18 at 5:49









              trancelocationtrancelocation

              10.8k1723




              10.8k1723












              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27




















              • $begingroup$
                how do i know $a neq 2$?
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:31










              • $begingroup$
                @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
                $endgroup$
                – Yadati Kiran
                Dec 12 '18 at 6:47










              • $begingroup$
                Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
                $endgroup$
                – stackofhay42
                Dec 12 '18 at 6:51












              • $begingroup$
                @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:25










              • $begingroup$
                @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
                $endgroup$
                – Martund
                Dec 12 '18 at 9:27


















              $begingroup$
              how do i know $a neq 2$?
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:31




              $begingroup$
              how do i know $a neq 2$?
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:31












              $begingroup$
              @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
              $endgroup$
              – Yadati Kiran
              Dec 12 '18 at 6:47




              $begingroup$
              @stackofhay42: If $a=2$, we t get the AM-GM $textbf{equality}$.
              $endgroup$
              – Yadati Kiran
              Dec 12 '18 at 6:47












              $begingroup$
              Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:51






              $begingroup$
              Yes, if $a = 2$, we get AM-GM equality, meaning that we need to show that we cannot have $a = 2$. We never showed that it cannot be true that $a = 2$. I think that this is circular reasoning.
              $endgroup$
              – stackofhay42
              Dec 12 '18 at 6:51














              $begingroup$
              @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:25




              $begingroup$
              @RakeshBhatt: By induction, we can say that all of the terms of the given sequence are positive. We know that $a_1>0$. Suppose $a_k>0 ,Longrightarrow a_{k+1}>0$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:25












              $begingroup$
              @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:27






              $begingroup$
              @stackofhay42: We can also prove that by induction. $a_1neq 2, a_kneq 2 Longrightarrow a_{k+1}neq 2$ by the given relation.
              $endgroup$
              – Martund
              Dec 12 '18 at 9:27













              2












              $begingroup$

              Note $a_n >0$, $n=1,2,3,.....$.



              Let $n ge 1$:



              $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



              $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



              $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



              (The first term $ge 0$, a square divided by a positive number.)



              Hence



              $a_{n+1} ge 2$, $n in mathbb{N}$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note $a_n >0$, $n=1,2,3,.....$.



                Let $n ge 1$:



                $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



                $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



                $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



                (The first term $ge 0$, a square divided by a positive number.)



                Hence



                $a_{n+1} ge 2$, $n in mathbb{N}$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note $a_n >0$, $n=1,2,3,.....$.



                  Let $n ge 1$:



                  $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



                  $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



                  $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



                  (The first term $ge 0$, a square divided by a positive number.)



                  Hence



                  $a_{n+1} ge 2$, $n in mathbb{N}$.






                  share|cite|improve this answer









                  $endgroup$



                  Note $a_n >0$, $n=1,2,3,.....$.



                  Let $n ge 1$:



                  $a_{n+1}= dfrac{a_n^2 +4}{2a_n}=$



                  $dfrac{(a_n-2)^2+4a_n}{2a_n}=$



                  $dfrac{(a_n-2)^2}{2a_n} + 2 ge 2$.



                  (The first term $ge 0$, a square divided by a positive number.)



                  Hence



                  $a_{n+1} ge 2$, $n in mathbb{N}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 11:01









                  Peter SzilasPeter Szilas

                  11.1k2821




                  11.1k2821























                      0












                      $begingroup$

                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12
















                      0












                      $begingroup$

                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12














                      0












                      0








                      0





                      $begingroup$

                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...






                      share|cite|improve this answer











                      $endgroup$



                      Following with your reasoning, suppose $x + 1/x = 2$. Then:
                      $$x + 1/x = 2implies x^2 - 2x + 1 = 0implies x = 1.$$
                      But this is impossible because...







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 12 '18 at 10:44

























                      answered Dec 12 '18 at 7:44









                      Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

                      34.3k42871




                      34.3k42871












                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12


















                      • $begingroup$
                        What about $a_n<2$?
                        $endgroup$
                        – Rakesh Bhatt
                        Dec 12 '18 at 10:14










                      • $begingroup$
                        @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 12 '18 at 10:43












                      • $begingroup$
                        Why is this impossible?
                        $endgroup$
                        – Ekesh Kumar
                        Dec 13 '18 at 23:45










                      • $begingroup$
                        @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                        $endgroup$
                        – Martín-Blas Pérez Pinilla
                        Dec 14 '18 at 7:12
















                      $begingroup$
                      What about $a_n<2$?
                      $endgroup$
                      – Rakesh Bhatt
                      Dec 12 '18 at 10:14




                      $begingroup$
                      What about $a_n<2$?
                      $endgroup$
                      – Rakesh Bhatt
                      Dec 12 '18 at 10:14












                      $begingroup$
                      @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 12 '18 at 10:43






                      $begingroup$
                      @RakeshBhatt, the OP has proved $a_n > 2implies a_{n+1}ge 2$. Only is required excluding the equality.
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 12 '18 at 10:43














                      $begingroup$
                      Why is this impossible?
                      $endgroup$
                      – Ekesh Kumar
                      Dec 13 '18 at 23:45




                      $begingroup$
                      Why is this impossible?
                      $endgroup$
                      – Ekesh Kumar
                      Dec 13 '18 at 23:45












                      $begingroup$
                      @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 14 '18 at 7:12




                      $begingroup$
                      @Ekesh, because you have supposed $a_n > 2$ ($x > 1$).
                      $endgroup$
                      – Martín-Blas Pérez Pinilla
                      Dec 14 '18 at 7:12


















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