$x,y in R$ and $x neq y$ Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$












0












$begingroup$


Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$



Struggling with this proof atm.



Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.






calculus inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is just the arithmetic-geometric mean inequality.
    $endgroup$
    – Wojowu
    Nov 29 '18 at 18:03










  • $begingroup$
    Can people please stop tagging real analysis to whatever has to do with calculus ?
    $endgroup$
    – Rebellos
    Nov 29 '18 at 18:05
















0












$begingroup$


Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$



Struggling with this proof atm.



Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.






calculus inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is just the arithmetic-geometric mean inequality.
    $endgroup$
    – Wojowu
    Nov 29 '18 at 18:03










  • $begingroup$
    Can people please stop tagging real analysis to whatever has to do with calculus ?
    $endgroup$
    – Rebellos
    Nov 29 '18 at 18:05














0












0








0





$begingroup$


Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$



Struggling with this proof atm.



Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.






calculus inequality







share|cite|improve this question











$endgroup$




Let $x,y in R$ and $x neq y$. Show that $e^frac{x+y}2 lt frac12 (e^x + e^y)$



Struggling with this proof atm.



Tried every approach I could think of. Attempted to look at it written as a series, as a limit and tried playing around with the general rules for exponential functions.






calculus inequality




calculus inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 18:23









Richard

3801211




3801211










asked Nov 29 '18 at 17:58









NhefNhef

91




91








  • 1




    $begingroup$
    This is just the arithmetic-geometric mean inequality.
    $endgroup$
    – Wojowu
    Nov 29 '18 at 18:03










  • $begingroup$
    Can people please stop tagging real analysis to whatever has to do with calculus ?
    $endgroup$
    – Rebellos
    Nov 29 '18 at 18:05














  • 1




    $begingroup$
    This is just the arithmetic-geometric mean inequality.
    $endgroup$
    – Wojowu
    Nov 29 '18 at 18:03










  • $begingroup$
    Can people please stop tagging real analysis to whatever has to do with calculus ?
    $endgroup$
    – Rebellos
    Nov 29 '18 at 18:05








1




1




$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03




$begingroup$
This is just the arithmetic-geometric mean inequality.
$endgroup$
– Wojowu
Nov 29 '18 at 18:03












$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05




$begingroup$
Can people please stop tagging real analysis to whatever has to do with calculus ?
$endgroup$
– Rebellos
Nov 29 '18 at 18:05










3 Answers
3






active

oldest

votes


















4












$begingroup$

if $ane b$



$(a-b)^2 > 0\
a^2 + b^2 > 2ab\
ab < frac 12 (a^2 + b^2)$



This is the AM-GM inequality.



$a = e^frac x2, b= e^frac y2$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$



    $$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$



    with $lambda=frac12$ we have



    $$e^{frac{x+y}2} < frac12 (e^x + e^y)$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        if $ane b$



        $(a-b)^2 > 0\
        a^2 + b^2 > 2ab\
        ab < frac 12 (a^2 + b^2)$



        This is the AM-GM inequality.



        $a = e^frac x2, b= e^frac y2$






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          if $ane b$



          $(a-b)^2 > 0\
          a^2 + b^2 > 2ab\
          ab < frac 12 (a^2 + b^2)$



          This is the AM-GM inequality.



          $a = e^frac x2, b= e^frac y2$






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            if $ane b$



            $(a-b)^2 > 0\
            a^2 + b^2 > 2ab\
            ab < frac 12 (a^2 + b^2)$



            This is the AM-GM inequality.



            $a = e^frac x2, b= e^frac y2$






            share|cite|improve this answer









            $endgroup$



            if $ane b$



            $(a-b)^2 > 0\
            a^2 + b^2 > 2ab\
            ab < frac 12 (a^2 + b^2)$



            This is the AM-GM inequality.



            $a = e^frac x2, b= e^frac y2$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 '18 at 18:03









            Doug MDoug M

            44.8k31854




            44.8k31854























                1












                $begingroup$

                By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$



                $$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$



                with $lambda=frac12$ we have



                $$e^{frac{x+y}2} < frac12 (e^x + e^y)$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$



                  $$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$



                  with $lambda=frac12$ we have



                  $$e^{frac{x+y}2} < frac12 (e^x + e^y)$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$



                    $$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$



                    with $lambda=frac12$ we have



                    $$e^{frac{x+y}2} < frac12 (e^x + e^y)$$






                    share|cite|improve this answer









                    $endgroup$



                    By convexity for $f(x)=e^x$, that is by Jensen's inequality for $xneq y$



                    $$f(lambda x+(1-lambda)y) < lambda f(x)+(1-lambda)f(y)$$



                    with $lambda=frac12$ we have



                    $$e^{frac{x+y}2} < frac12 (e^x + e^y)$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 29 '18 at 18:16









                    gimusigimusi

                    92.8k84494




                    92.8k84494























                        0












                        $begingroup$

                        We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$






                            share|cite|improve this answer









                            $endgroup$



                            We have $${(e^{xover 2}-e^{yover 2})^2>0\e^x+e^y>2e^{xover 2}e^{yover 2}\{1over 2}(e^x+e^y)>e^{x+yover 2}}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 29 '18 at 18:21









                            Mostafa AyazMostafa Ayaz

                            15.5k3939




                            15.5k3939






























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