Is there a name for this inequality?

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6
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Consider two positive real numbers $frac1a$ and $frac1b$. Show
that the 'mean' $$dfrac1fraca+b2=dfrac2a+b$$ will
always be less than the arithmetic mean $$dfraca+b2ab.$$




Proving this is not difficult, but I was wondering if there was a name for this inequality, or if it demonstrates another known inequality in action, much like how many known inequalities are simple consequences of say the AM/GM Inequality.




inequality terminology




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  • 1




    It is probably more insightful if you replace $frac 1a$ and $frac 1b$ by $a$ and $b$.
    – Arnaud Mortier
    Aug 24 at 6:55










  • @ArnaudMortier If we looked at $a$ instead of $1/a$, we would explicitly have to include a non-$0$ criterion. There’s a tradeoff.
    – Chase Ryan Taylor
    Aug 24 at 7:18






  • 3




    @ChaseRyanTaylor since the numbers have to be positive either way, that's not an issue, at least in English.
    – Especially Lime
    Aug 24 at 7:29














up vote
6
down vote

favorite













Consider two positive real numbers $frac1a$ and $frac1b$. Show
that the 'mean' $$dfrac1fraca+b2=dfrac2a+b$$ will
always be less than the arithmetic mean $$dfraca+b2ab.$$




Proving this is not difficult, but I was wondering if there was a name for this inequality, or if it demonstrates another known inequality in action, much like how many known inequalities are simple consequences of say the AM/GM Inequality.




inequality terminology




share|cite|improve this question


















  • 1




    It is probably more insightful if you replace $frac 1a$ and $frac 1b$ by $a$ and $b$.
    – Arnaud Mortier
    Aug 24 at 6:55










  • @ArnaudMortier If we looked at $a$ instead of $1/a$, we would explicitly have to include a non-$0$ criterion. There’s a tradeoff.
    – Chase Ryan Taylor
    Aug 24 at 7:18






  • 3




    @ChaseRyanTaylor since the numbers have to be positive either way, that's not an issue, at least in English.
    – Especially Lime
    Aug 24 at 7:29












up vote
6
down vote

favorite









up vote
6
down vote

favorite












Consider two positive real numbers $frac1a$ and $frac1b$. Show
that the 'mean' $$dfrac1fraca+b2=dfrac2a+b$$ will
always be less than the arithmetic mean $$dfraca+b2ab.$$




Proving this is not difficult, but I was wondering if there was a name for this inequality, or if it demonstrates another known inequality in action, much like how many known inequalities are simple consequences of say the AM/GM Inequality.




inequality terminology




share|cite|improve this question















Consider two positive real numbers $frac1a$ and $frac1b$. Show
that the 'mean' $$dfrac1fraca+b2=dfrac2a+b$$ will
always be less than the arithmetic mean $$dfraca+b2ab.$$




Proving this is not difficult, but I was wondering if there was a name for this inequality, or if it demonstrates another known inequality in action, much like how many known inequalities are simple consequences of say the AM/GM Inequality.





inequality terminology





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 7:29









Especially Lime

19.3k22252




19.3k22252










asked Aug 24 at 6:47









Trogdor

5,03361741




5,03361741







  • 1




    It is probably more insightful if you replace $frac 1a$ and $frac 1b$ by $a$ and $b$.
    – Arnaud Mortier
    Aug 24 at 6:55










  • @ArnaudMortier If we looked at $a$ instead of $1/a$, we would explicitly have to include a non-$0$ criterion. There’s a tradeoff.
    – Chase Ryan Taylor
    Aug 24 at 7:18






  • 3




    @ChaseRyanTaylor since the numbers have to be positive either way, that's not an issue, at least in English.
    – Especially Lime
    Aug 24 at 7:29












  • 1




    It is probably more insightful if you replace $frac 1a$ and $frac 1b$ by $a$ and $b$.
    – Arnaud Mortier
    Aug 24 at 6:55










  • @ArnaudMortier If we looked at $a$ instead of $1/a$, we would explicitly have to include a non-$0$ criterion. There’s a tradeoff.
    – Chase Ryan Taylor
    Aug 24 at 7:18






  • 3




    @ChaseRyanTaylor since the numbers have to be positive either way, that's not an issue, at least in English.
    – Especially Lime
    Aug 24 at 7:29







1




1




It is probably more insightful if you replace $frac 1a$ and $frac 1b$ by $a$ and $b$.
– Arnaud Mortier
Aug 24 at 6:55




It is probably more insightful if you replace $frac 1a$ and $frac 1b$ by $a$ and $b$.
– Arnaud Mortier
Aug 24 at 6:55












@ArnaudMortier If we looked at $a$ instead of $1/a$, we would explicitly have to include a non-$0$ criterion. There’s a tradeoff.
– Chase Ryan Taylor
Aug 24 at 7:18




@ArnaudMortier If we looked at $a$ instead of $1/a$, we would explicitly have to include a non-$0$ criterion. There’s a tradeoff.
– Chase Ryan Taylor
Aug 24 at 7:18




3




3




@ChaseRyanTaylor since the numbers have to be positive either way, that's not an issue, at least in English.
– Especially Lime
Aug 24 at 7:29




@ChaseRyanTaylor since the numbers have to be positive either way, that's not an issue, at least in English.
– Especially Lime
Aug 24 at 7:29










2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










Yes this is the HM-AM inequality and it is a consequence of AM-GM.



Take a look here



RMS-AM-GM-HM inequalities






share|cite|improve this answer
















  • 1




    HM is for Harmonic Mean.
    – mathreadler
    Aug 24 at 7:07










  • @mathreadler Yes exactly!
    – gimusi
    Aug 24 at 7:20

















up vote
2
down vote













This is just saying that the inverse of the mean is at most the mean of the inverses:



$$frac1fraca+b2 leq fracfrac1a + frac1b 2$$



This follows from the convexity of the function $frac 1 x$ (on the positive real numbers). Of course, convexity is closely related to the AM-GM inequality, so maybe it's not surprising that this can also been seen as related to AM-GM.






share|cite|improve this answer




















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    Yes this is the HM-AM inequality and it is a consequence of AM-GM.



    Take a look here



    RMS-AM-GM-HM inequalities






    share|cite|improve this answer
















    • 1




      HM is for Harmonic Mean.
      – mathreadler
      Aug 24 at 7:07










    • @mathreadler Yes exactly!
      – gimusi
      Aug 24 at 7:20














    up vote
    5
    down vote



    accepted










    Yes this is the HM-AM inequality and it is a consequence of AM-GM.



    Take a look here



    RMS-AM-GM-HM inequalities






    share|cite|improve this answer
















    • 1




      HM is for Harmonic Mean.
      – mathreadler
      Aug 24 at 7:07










    • @mathreadler Yes exactly!
      – gimusi
      Aug 24 at 7:20












    up vote
    5
    down vote



    accepted







    up vote
    5
    down vote



    accepted






    Yes this is the HM-AM inequality and it is a consequence of AM-GM.



    Take a look here



    RMS-AM-GM-HM inequalities






    share|cite|improve this answer












    Yes this is the HM-AM inequality and it is a consequence of AM-GM.



    Take a look here



    RMS-AM-GM-HM inequalities







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 24 at 6:52









    gimusi

    69.7k73686




    69.7k73686







    • 1




      HM is for Harmonic Mean.
      – mathreadler
      Aug 24 at 7:07










    • @mathreadler Yes exactly!
      – gimusi
      Aug 24 at 7:20












    • 1




      HM is for Harmonic Mean.
      – mathreadler
      Aug 24 at 7:07










    • @mathreadler Yes exactly!
      – gimusi
      Aug 24 at 7:20







    1




    1




    HM is for Harmonic Mean.
    – mathreadler
    Aug 24 at 7:07




    HM is for Harmonic Mean.
    – mathreadler
    Aug 24 at 7:07












    @mathreadler Yes exactly!
    – gimusi
    Aug 24 at 7:20




    @mathreadler Yes exactly!
    – gimusi
    Aug 24 at 7:20










    up vote
    2
    down vote













    This is just saying that the inverse of the mean is at most the mean of the inverses:



    $$frac1fraca+b2 leq fracfrac1a + frac1b 2$$



    This follows from the convexity of the function $frac 1 x$ (on the positive real numbers). Of course, convexity is closely related to the AM-GM inequality, so maybe it's not surprising that this can also been seen as related to AM-GM.






    share|cite|improve this answer
























      up vote
      2
      down vote













      This is just saying that the inverse of the mean is at most the mean of the inverses:



      $$frac1fraca+b2 leq fracfrac1a + frac1b 2$$



      This follows from the convexity of the function $frac 1 x$ (on the positive real numbers). Of course, convexity is closely related to the AM-GM inequality, so maybe it's not surprising that this can also been seen as related to AM-GM.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        This is just saying that the inverse of the mean is at most the mean of the inverses:



        $$frac1fraca+b2 leq fracfrac1a + frac1b 2$$



        This follows from the convexity of the function $frac 1 x$ (on the positive real numbers). Of course, convexity is closely related to the AM-GM inequality, so maybe it's not surprising that this can also been seen as related to AM-GM.






        share|cite|improve this answer












        This is just saying that the inverse of the mean is at most the mean of the inverses:



        $$frac1fraca+b2 leq fracfrac1a + frac1b 2$$



        This follows from the convexity of the function $frac 1 x$ (on the positive real numbers). Of course, convexity is closely related to the AM-GM inequality, so maybe it's not surprising that this can also been seen as related to AM-GM.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 24 at 7:23









        Jack M

        17.4k33473




        17.4k33473



























             

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