Finding bounds of the set: $ A = { frac{mcdot n}{m+n}: m,n in mathbb N } $











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I'm having some problems with this.



I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.



Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.



Thank you for any hints.






upper-lower-bounds







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    up vote
    1
    down vote

    favorite












    I'm having some problems with this.



    I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.



    Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.



    Thank you for any hints.






    upper-lower-bounds







    share|cite|improve this question









    New contributor




    Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm having some problems with this.



      I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.



      Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.



      Thank you for any hints.






      upper-lower-bounds







      share|cite|improve this question









      New contributor




      Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm having some problems with this.



      I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.



      Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.



      Thank you for any hints.






      upper-lower-bounds




      upper-lower-bounds






      share|cite|improve this question









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      Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      edited yesterday









      amWhy

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          3 Answers
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          active

          oldest

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          up vote
          2
          down vote



          accepted










          The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.



          On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.






          share|cite|improve this answer




























            up vote
            1
            down vote













            Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.






            share|cite|improve this answer




























              up vote
              0
              down vote













              Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.



              Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?



              Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?



              If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.






              share|cite|improve this answer





















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                3 Answers
                3






                active

                oldest

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                3 Answers
                3






                active

                oldest

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                active

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                active

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                up vote
                2
                down vote



                accepted










                The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.



                On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.






                share|cite|improve this answer

























                  up vote
                  2
                  down vote



                  accepted










                  The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.



                  On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.






                  share|cite|improve this answer























                    up vote
                    2
                    down vote



                    accepted







                    up vote
                    2
                    down vote



                    accepted






                    The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.



                    On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.






                    share|cite|improve this answer












                    The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.



                    On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    José Carlos Santos

                    139k18111203




                    139k18111203






















                        up vote
                        1
                        down vote













                        Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.






                            share|cite|improve this answer












                            Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            Richard Martin

                            1,3238




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                                up vote
                                0
                                down vote













                                Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.



                                Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?



                                Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?



                                If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.






                                share|cite|improve this answer

























                                  up vote
                                  0
                                  down vote













                                  Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.



                                  Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?



                                  Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?



                                  If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.






                                  share|cite|improve this answer























                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.



                                    Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?



                                    Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?



                                    If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.






                                    share|cite|improve this answer












                                    Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.



                                    Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?



                                    Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?



                                    If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered yesterday









                                    Ovi

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