Show inequality holds by induction












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$begingroup$


Say I have $x_1 > y_1 > 0$ and $x_{n+1} = frac{x_n + y_n}{2}, y_{n+1} = frac{2x_ny_n}{x_n + y_n}$ for $n geq 1$. Show that $x_n > x_{n+1} > y_{n+1} > y_n > 0$.



So the obvious approach to me it seems is proceed by induction.



So with our base case, $n = 1$, $x_1 > frac{x_1 + y_1}{2} > frac{2x_1y_1}{x_1 + y_1} > y_1 > 0$.



Not sure if there's a good way to clean this up but not even sure how to show this holds true for the base case.






real-analysis sequences-and-series induction







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    $begingroup$


    Say I have $x_1 > y_1 > 0$ and $x_{n+1} = frac{x_n + y_n}{2}, y_{n+1} = frac{2x_ny_n}{x_n + y_n}$ for $n geq 1$. Show that $x_n > x_{n+1} > y_{n+1} > y_n > 0$.



    So the obvious approach to me it seems is proceed by induction.



    So with our base case, $n = 1$, $x_1 > frac{x_1 + y_1}{2} > frac{2x_1y_1}{x_1 + y_1} > y_1 > 0$.



    Not sure if there's a good way to clean this up but not even sure how to show this holds true for the base case.






    real-analysis sequences-and-series induction







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Say I have $x_1 > y_1 > 0$ and $x_{n+1} = frac{x_n + y_n}{2}, y_{n+1} = frac{2x_ny_n}{x_n + y_n}$ for $n geq 1$. Show that $x_n > x_{n+1} > y_{n+1} > y_n > 0$.



      So the obvious approach to me it seems is proceed by induction.



      So with our base case, $n = 1$, $x_1 > frac{x_1 + y_1}{2} > frac{2x_1y_1}{x_1 + y_1} > y_1 > 0$.



      Not sure if there's a good way to clean this up but not even sure how to show this holds true for the base case.






      real-analysis sequences-and-series induction







      share|cite|improve this question









      $endgroup$




      Say I have $x_1 > y_1 > 0$ and $x_{n+1} = frac{x_n + y_n}{2}, y_{n+1} = frac{2x_ny_n}{x_n + y_n}$ for $n geq 1$. Show that $x_n > x_{n+1} > y_{n+1} > y_n > 0$.



      So the obvious approach to me it seems is proceed by induction.



      So with our base case, $n = 1$, $x_1 > frac{x_1 + y_1}{2} > frac{2x_1y_1}{x_1 + y_1} > y_1 > 0$.



      Not sure if there's a good way to clean this up but not even sure how to show this holds true for the base case.






      real-analysis sequences-and-series induction




      real-analysis sequences-and-series induction






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      asked Sep 28 '18 at 1:09









      SS'SS'

      597314




      597314






















          3 Answers
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          $begingroup$

          Your inductive hypothesis is $x_n > y_n$ then there are $3$ things to show firstly
          begin{eqnarray*}
          x_n = frac{x_n+x_n}{2} > frac{x_n+y_n}{2} > x_{n+1}.
          end{eqnarray*}
          Secondly AM-HM which follows from $(x_n-y_n)^2>0$
          begin{eqnarray*}
          frac{x_n+y_n}{2} > frac{2x_n y_n}{x_n+y_n}.
          end{eqnarray*}
          Thirdly we have $x_n y_n > y_n^2$ now add $x_n y_n$ to both sides and divide by $x_n+y_n$ and we have
          begin{eqnarray*}
          y_{n+1}= frac{2x_n y_n}{x_n+y_n}> y_n.
          end{eqnarray*}






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            We don’t need induction indeed we have that



            $$x_n > x_{n+1}iff x_n =frac{x_n + x_n}{2}>frac{x_n + y_n}{2}$$



            $$x_{n+1} >y_{n+1}iff frac{x_n + y_n}{2}> frac{2x_ny_n}{x_n + y_n}iff (x_n-y_n)^2>0$$



            $$Y_{n+1} >y_{n}iff frac{2x_ny_n}{x_n + y_n}>y_niff frac{x_n+x_n}{x_n + y_n}>1$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Notice that for all $n$ we have $x_n,y_n>0$. Therefore if $$x_n>y_n>0$$ we have $$(x_n-y_n)^2>0\(x_n+y_n)^2>4x_ny_n\{x_n+y_nover 2}>{2x_ny_nover x_n+y_n}\x_{n+1}>y_{n+1}$$since $x_1>y_1>0$ we have $x_n>y_n>0$ for all $n$. Also $$x_{n+1}={x_n+y_nover 2}<{x_n+x_nover 2}=x_n\y_{n+1}={2x_ny_nover x_n+y_n}$$since ${2auover a+u}$ is an increasing function of $u$ for $a>0$ and $uge a$, the minimum is attained when $u=a$. Therefore $$y_{n+1}={2x_ny_nover x_n+y_n}>{2y_ny_nover y_n+y_n}=y_n$$from which we finally conclude that$$x_{n}>x_{n+1}>y_{n+1}>y_{n}$$






              share|cite|improve this answer









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                3 Answers
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                active

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                3 Answers
                3






                active

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                active

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                active

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                0












                $begingroup$

                Your inductive hypothesis is $x_n > y_n$ then there are $3$ things to show firstly
                begin{eqnarray*}
                x_n = frac{x_n+x_n}{2} > frac{x_n+y_n}{2} > x_{n+1}.
                end{eqnarray*}
                Secondly AM-HM which follows from $(x_n-y_n)^2>0$
                begin{eqnarray*}
                frac{x_n+y_n}{2} > frac{2x_n y_n}{x_n+y_n}.
                end{eqnarray*}
                Thirdly we have $x_n y_n > y_n^2$ now add $x_n y_n$ to both sides and divide by $x_n+y_n$ and we have
                begin{eqnarray*}
                y_{n+1}= frac{2x_n y_n}{x_n+y_n}> y_n.
                end{eqnarray*}






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Your inductive hypothesis is $x_n > y_n$ then there are $3$ things to show firstly
                  begin{eqnarray*}
                  x_n = frac{x_n+x_n}{2} > frac{x_n+y_n}{2} > x_{n+1}.
                  end{eqnarray*}
                  Secondly AM-HM which follows from $(x_n-y_n)^2>0$
                  begin{eqnarray*}
                  frac{x_n+y_n}{2} > frac{2x_n y_n}{x_n+y_n}.
                  end{eqnarray*}
                  Thirdly we have $x_n y_n > y_n^2$ now add $x_n y_n$ to both sides and divide by $x_n+y_n$ and we have
                  begin{eqnarray*}
                  y_{n+1}= frac{2x_n y_n}{x_n+y_n}> y_n.
                  end{eqnarray*}






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Your inductive hypothesis is $x_n > y_n$ then there are $3$ things to show firstly
                    begin{eqnarray*}
                    x_n = frac{x_n+x_n}{2} > frac{x_n+y_n}{2} > x_{n+1}.
                    end{eqnarray*}
                    Secondly AM-HM which follows from $(x_n-y_n)^2>0$
                    begin{eqnarray*}
                    frac{x_n+y_n}{2} > frac{2x_n y_n}{x_n+y_n}.
                    end{eqnarray*}
                    Thirdly we have $x_n y_n > y_n^2$ now add $x_n y_n$ to both sides and divide by $x_n+y_n$ and we have
                    begin{eqnarray*}
                    y_{n+1}= frac{2x_n y_n}{x_n+y_n}> y_n.
                    end{eqnarray*}






                    share|cite|improve this answer









                    $endgroup$



                    Your inductive hypothesis is $x_n > y_n$ then there are $3$ things to show firstly
                    begin{eqnarray*}
                    x_n = frac{x_n+x_n}{2} > frac{x_n+y_n}{2} > x_{n+1}.
                    end{eqnarray*}
                    Secondly AM-HM which follows from $(x_n-y_n)^2>0$
                    begin{eqnarray*}
                    frac{x_n+y_n}{2} > frac{2x_n y_n}{x_n+y_n}.
                    end{eqnarray*}
                    Thirdly we have $x_n y_n > y_n^2$ now add $x_n y_n$ to both sides and divide by $x_n+y_n$ and we have
                    begin{eqnarray*}
                    y_{n+1}= frac{2x_n y_n}{x_n+y_n}> y_n.
                    end{eqnarray*}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 28 '18 at 1:25









                    Donald SplutterwitDonald Splutterwit

                    22.6k21446




                    22.6k21446























                        0












                        $begingroup$

                        We don’t need induction indeed we have that



                        $$x_n > x_{n+1}iff x_n =frac{x_n + x_n}{2}>frac{x_n + y_n}{2}$$



                        $$x_{n+1} >y_{n+1}iff frac{x_n + y_n}{2}> frac{2x_ny_n}{x_n + y_n}iff (x_n-y_n)^2>0$$



                        $$Y_{n+1} >y_{n}iff frac{2x_ny_n}{x_n + y_n}>y_niff frac{x_n+x_n}{x_n + y_n}>1$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          We don’t need induction indeed we have that



                          $$x_n > x_{n+1}iff x_n =frac{x_n + x_n}{2}>frac{x_n + y_n}{2}$$



                          $$x_{n+1} >y_{n+1}iff frac{x_n + y_n}{2}> frac{2x_ny_n}{x_n + y_n}iff (x_n-y_n)^2>0$$



                          $$Y_{n+1} >y_{n}iff frac{2x_ny_n}{x_n + y_n}>y_niff frac{x_n+x_n}{x_n + y_n}>1$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            We don’t need induction indeed we have that



                            $$x_n > x_{n+1}iff x_n =frac{x_n + x_n}{2}>frac{x_n + y_n}{2}$$



                            $$x_{n+1} >y_{n+1}iff frac{x_n + y_n}{2}> frac{2x_ny_n}{x_n + y_n}iff (x_n-y_n)^2>0$$



                            $$Y_{n+1} >y_{n}iff frac{2x_ny_n}{x_n + y_n}>y_niff frac{x_n+x_n}{x_n + y_n}>1$$






                            share|cite|improve this answer









                            $endgroup$



                            We don’t need induction indeed we have that



                            $$x_n > x_{n+1}iff x_n =frac{x_n + x_n}{2}>frac{x_n + y_n}{2}$$



                            $$x_{n+1} >y_{n+1}iff frac{x_n + y_n}{2}> frac{2x_ny_n}{x_n + y_n}iff (x_n-y_n)^2>0$$



                            $$Y_{n+1} >y_{n}iff frac{2x_ny_n}{x_n + y_n}>y_niff frac{x_n+x_n}{x_n + y_n}>1$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 28 '18 at 1:24









                            gimusigimusi

                            92.9k94494




                            92.9k94494























                                0












                                $begingroup$

                                Notice that for all $n$ we have $x_n,y_n>0$. Therefore if $$x_n>y_n>0$$ we have $$(x_n-y_n)^2>0\(x_n+y_n)^2>4x_ny_n\{x_n+y_nover 2}>{2x_ny_nover x_n+y_n}\x_{n+1}>y_{n+1}$$since $x_1>y_1>0$ we have $x_n>y_n>0$ for all $n$. Also $$x_{n+1}={x_n+y_nover 2}<{x_n+x_nover 2}=x_n\y_{n+1}={2x_ny_nover x_n+y_n}$$since ${2auover a+u}$ is an increasing function of $u$ for $a>0$ and $uge a$, the minimum is attained when $u=a$. Therefore $$y_{n+1}={2x_ny_nover x_n+y_n}>{2y_ny_nover y_n+y_n}=y_n$$from which we finally conclude that$$x_{n}>x_{n+1}>y_{n+1}>y_{n}$$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Notice that for all $n$ we have $x_n,y_n>0$. Therefore if $$x_n>y_n>0$$ we have $$(x_n-y_n)^2>0\(x_n+y_n)^2>4x_ny_n\{x_n+y_nover 2}>{2x_ny_nover x_n+y_n}\x_{n+1}>y_{n+1}$$since $x_1>y_1>0$ we have $x_n>y_n>0$ for all $n$. Also $$x_{n+1}={x_n+y_nover 2}<{x_n+x_nover 2}=x_n\y_{n+1}={2x_ny_nover x_n+y_n}$$since ${2auover a+u}$ is an increasing function of $u$ for $a>0$ and $uge a$, the minimum is attained when $u=a$. Therefore $$y_{n+1}={2x_ny_nover x_n+y_n}>{2y_ny_nover y_n+y_n}=y_n$$from which we finally conclude that$$x_{n}>x_{n+1}>y_{n+1}>y_{n}$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Notice that for all $n$ we have $x_n,y_n>0$. Therefore if $$x_n>y_n>0$$ we have $$(x_n-y_n)^2>0\(x_n+y_n)^2>4x_ny_n\{x_n+y_nover 2}>{2x_ny_nover x_n+y_n}\x_{n+1}>y_{n+1}$$since $x_1>y_1>0$ we have $x_n>y_n>0$ for all $n$. Also $$x_{n+1}={x_n+y_nover 2}<{x_n+x_nover 2}=x_n\y_{n+1}={2x_ny_nover x_n+y_n}$$since ${2auover a+u}$ is an increasing function of $u$ for $a>0$ and $uge a$, the minimum is attained when $u=a$. Therefore $$y_{n+1}={2x_ny_nover x_n+y_n}>{2y_ny_nover y_n+y_n}=y_n$$from which we finally conclude that$$x_{n}>x_{n+1}>y_{n+1}>y_{n}$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Notice that for all $n$ we have $x_n,y_n>0$. Therefore if $$x_n>y_n>0$$ we have $$(x_n-y_n)^2>0\(x_n+y_n)^2>4x_ny_n\{x_n+y_nover 2}>{2x_ny_nover x_n+y_n}\x_{n+1}>y_{n+1}$$since $x_1>y_1>0$ we have $x_n>y_n>0$ for all $n$. Also $$x_{n+1}={x_n+y_nover 2}<{x_n+x_nover 2}=x_n\y_{n+1}={2x_ny_nover x_n+y_n}$$since ${2auover a+u}$ is an increasing function of $u$ for $a>0$ and $uge a$, the minimum is attained when $u=a$. Therefore $$y_{n+1}={2x_ny_nover x_n+y_n}>{2y_ny_nover y_n+y_n}=y_n$$from which we finally conclude that$$x_{n}>x_{n+1}>y_{n+1}>y_{n}$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 1 '18 at 9:00









                                    Mostafa AyazMostafa Ayaz

                                    15.4k3939




                                    15.4k3939






























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