Further terms of Taylor expansion for improved inequalities?












0












$begingroup$


Background: We know the classical exponential inequality
begin{align*}
1 + x le e^x qquad text{or} qquad x le e^{x - 1}
end{align*}
Taking $x_i = y_i/overline{y}$, where $y_i ge 0$ and $overline{y}$ is the arithmetic mean, and multiplying out the second form, we yield
begin{align*}
frac{prod_{i=1}^{n}y_i}{overline{y}^n} le expleft(frac{sum_{i=1}^{n}y_i}{overline{y}} - nright) = 1
end{align*}
And this yields a proof of the arithmetic-means and geometric-means (AM-GM) inequality.



An improved exponential inequality: We also have the exponential inequality
begin{align*}
1 + x + frac{x^2}{2} + frac{x^3}{6} le e^x
end{align*}
If we substitute $y = 1 + x + frac{x^2}{2} + frac{x^3}{6}$, we get the equivalent form
begin{align*}
y le e^{g(y)}
end{align*}
where
begin{align*}
g(y) = left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^frac{1}{3} - left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^{-frac{1}{3}} - 1
end{align*}
is the inverse function expressing $x$ in terms of $y$.



Can we derive other inequalities? From here, there are many directions we may take with our given inequality, but due to the non-linear nature of $g(y)$, a direct trick of normalization as in the AM-GM proof will yield no immediate discoveries.



This post is meant to be very open to discuss possible inequality sharpenings from the cubic-Taylor expansion inequality above. Some observations:




  1. $(g(y) + 1)^3 = -3(g(y) + 1)$


  2. $[6y-2]_+^frac{1}{3} - [6y-2]_+^{-frac{1}{3}} - 1 < g(y) le ([6y-2]_+ + 1)^frac{1}{3} - ([6y-2]_+ + 1)^{-frac{1}{3}} - 1$, where $[x]_+ = max(x, 0)$.


Have fun!






algebra-precalculus inequality recreational-mathematics a.m.-g.m.-inequality







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Background: We know the classical exponential inequality
    begin{align*}
    1 + x le e^x qquad text{or} qquad x le e^{x - 1}
    end{align*}
    Taking $x_i = y_i/overline{y}$, where $y_i ge 0$ and $overline{y}$ is the arithmetic mean, and multiplying out the second form, we yield
    begin{align*}
    frac{prod_{i=1}^{n}y_i}{overline{y}^n} le expleft(frac{sum_{i=1}^{n}y_i}{overline{y}} - nright) = 1
    end{align*}
    And this yields a proof of the arithmetic-means and geometric-means (AM-GM) inequality.



    An improved exponential inequality: We also have the exponential inequality
    begin{align*}
    1 + x + frac{x^2}{2} + frac{x^3}{6} le e^x
    end{align*}
    If we substitute $y = 1 + x + frac{x^2}{2} + frac{x^3}{6}$, we get the equivalent form
    begin{align*}
    y le e^{g(y)}
    end{align*}
    where
    begin{align*}
    g(y) = left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^frac{1}{3} - left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^{-frac{1}{3}} - 1
    end{align*}
    is the inverse function expressing $x$ in terms of $y$.



    Can we derive other inequalities? From here, there are many directions we may take with our given inequality, but due to the non-linear nature of $g(y)$, a direct trick of normalization as in the AM-GM proof will yield no immediate discoveries.



    This post is meant to be very open to discuss possible inequality sharpenings from the cubic-Taylor expansion inequality above. Some observations:




    1. $(g(y) + 1)^3 = -3(g(y) + 1)$


    2. $[6y-2]_+^frac{1}{3} - [6y-2]_+^{-frac{1}{3}} - 1 < g(y) le ([6y-2]_+ + 1)^frac{1}{3} - ([6y-2]_+ + 1)^{-frac{1}{3}} - 1$, where $[x]_+ = max(x, 0)$.


    Have fun!






    algebra-precalculus inequality recreational-mathematics a.m.-g.m.-inequality







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Background: We know the classical exponential inequality
      begin{align*}
      1 + x le e^x qquad text{or} qquad x le e^{x - 1}
      end{align*}
      Taking $x_i = y_i/overline{y}$, where $y_i ge 0$ and $overline{y}$ is the arithmetic mean, and multiplying out the second form, we yield
      begin{align*}
      frac{prod_{i=1}^{n}y_i}{overline{y}^n} le expleft(frac{sum_{i=1}^{n}y_i}{overline{y}} - nright) = 1
      end{align*}
      And this yields a proof of the arithmetic-means and geometric-means (AM-GM) inequality.



      An improved exponential inequality: We also have the exponential inequality
      begin{align*}
      1 + x + frac{x^2}{2} + frac{x^3}{6} le e^x
      end{align*}
      If we substitute $y = 1 + x + frac{x^2}{2} + frac{x^3}{6}$, we get the equivalent form
      begin{align*}
      y le e^{g(y)}
      end{align*}
      where
      begin{align*}
      g(y) = left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^frac{1}{3} - left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^{-frac{1}{3}} - 1
      end{align*}
      is the inverse function expressing $x$ in terms of $y$.



      Can we derive other inequalities? From here, there are many directions we may take with our given inequality, but due to the non-linear nature of $g(y)$, a direct trick of normalization as in the AM-GM proof will yield no immediate discoveries.



      This post is meant to be very open to discuss possible inequality sharpenings from the cubic-Taylor expansion inequality above. Some observations:




      1. $(g(y) + 1)^3 = -3(g(y) + 1)$


      2. $[6y-2]_+^frac{1}{3} - [6y-2]_+^{-frac{1}{3}} - 1 < g(y) le ([6y-2]_+ + 1)^frac{1}{3} - ([6y-2]_+ + 1)^{-frac{1}{3}} - 1$, where $[x]_+ = max(x, 0)$.


      Have fun!






      algebra-precalculus inequality recreational-mathematics a.m.-g.m.-inequality







      share|cite|improve this question









      $endgroup$




      Background: We know the classical exponential inequality
      begin{align*}
      1 + x le e^x qquad text{or} qquad x le e^{x - 1}
      end{align*}
      Taking $x_i = y_i/overline{y}$, where $y_i ge 0$ and $overline{y}$ is the arithmetic mean, and multiplying out the second form, we yield
      begin{align*}
      frac{prod_{i=1}^{n}y_i}{overline{y}^n} le expleft(frac{sum_{i=1}^{n}y_i}{overline{y}} - nright) = 1
      end{align*}
      And this yields a proof of the arithmetic-means and geometric-means (AM-GM) inequality.



      An improved exponential inequality: We also have the exponential inequality
      begin{align*}
      1 + x + frac{x^2}{2} + frac{x^3}{6} le e^x
      end{align*}
      If we substitute $y = 1 + x + frac{x^2}{2} + frac{x^3}{6}$, we get the equivalent form
      begin{align*}
      y le e^{g(y)}
      end{align*}
      where
      begin{align*}
      g(y) = left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^frac{1}{3} - left(sqrt{9y^2 - 6y + 2} + 3y - 1right)^{-frac{1}{3}} - 1
      end{align*}
      is the inverse function expressing $x$ in terms of $y$.



      Can we derive other inequalities? From here, there are many directions we may take with our given inequality, but due to the non-linear nature of $g(y)$, a direct trick of normalization as in the AM-GM proof will yield no immediate discoveries.



      This post is meant to be very open to discuss possible inequality sharpenings from the cubic-Taylor expansion inequality above. Some observations:




      1. $(g(y) + 1)^3 = -3(g(y) + 1)$


      2. $[6y-2]_+^frac{1}{3} - [6y-2]_+^{-frac{1}{3}} - 1 < g(y) le ([6y-2]_+ + 1)^frac{1}{3} - ([6y-2]_+ + 1)^{-frac{1}{3}} - 1$, where $[x]_+ = max(x, 0)$.


      Have fun!






      algebra-precalculus inequality recreational-mathematics a.m.-g.m.-inequality




      algebra-precalculus inequality recreational-mathematics a.m.-g.m.-inequality






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      asked Dec 25 '18 at 8:08









      Tom ChenTom Chen

      2,163715




      2,163715






















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          $begingroup$

          By your idea we can get for example the following inequality.




          Let $a$, $b$ and $c$ be positive numbers such that $abc=1$.



          Prove that:
          $$(a+b)^{a+b}+(b+c)^{b+c}+(c+a)^{c+a}geq4(a+b+c).$$







          share|cite|improve this answer











          $endgroup$














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            $begingroup$

            By your idea we can get for example the following inequality.




            Let $a$, $b$ and $c$ be positive numbers such that $abc=1$.



            Prove that:
            $$(a+b)^{a+b}+(b+c)^{b+c}+(c+a)^{c+a}geq4(a+b+c).$$







            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              By your idea we can get for example the following inequality.




              Let $a$, $b$ and $c$ be positive numbers such that $abc=1$.



              Prove that:
              $$(a+b)^{a+b}+(b+c)^{b+c}+(c+a)^{c+a}geq4(a+b+c).$$







              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                By your idea we can get for example the following inequality.




                Let $a$, $b$ and $c$ be positive numbers such that $abc=1$.



                Prove that:
                $$(a+b)^{a+b}+(b+c)^{b+c}+(c+a)^{c+a}geq4(a+b+c).$$







                share|cite|improve this answer











                $endgroup$



                By your idea we can get for example the following inequality.




                Let $a$, $b$ and $c$ be positive numbers such that $abc=1$.



                Prove that:
                $$(a+b)^{a+b}+(b+c)^{b+c}+(c+a)^{c+a}geq4(a+b+c).$$








                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 25 '18 at 8:50

























                answered Dec 25 '18 at 8:39









                Michael RozenbergMichael Rozenberg

                111k1897201




                111k1897201






























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