Proving $b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$ where $0 leq b leq a leq 1$












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I am trying to prove an inequality which boils down to the following:
$$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$



Another equivalent formulation is:
$$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$



How can I prove it?






inequality logarithms exponentiation







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    0












    $begingroup$


    I am trying to prove an inequality which boils down to the following:
    $$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$



    Another equivalent formulation is:
    $$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$



    How can I prove it?






    inequality logarithms exponentiation







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to prove an inequality which boils down to the following:
      $$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$



      Another equivalent formulation is:
      $$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$



      How can I prove it?






      inequality logarithms exponentiation







      share|cite|improve this question









      $endgroup$




      I am trying to prove an inequality which boils down to the following:
      $$b^a(1-b)^{1-a} leq a^a(1-a)^{1-a}$$ where $0 leq b leq a leq 1$



      Another equivalent formulation is:
      $$ alog{b} + (1-a)log{(1-b)} leq alog a + (1-a)log (1-a)$$



      How can I prove it?






      inequality logarithms exponentiation




      inequality logarithms exponentiation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 29 at 4:51









      hi15hi15

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      1586






















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          $begingroup$

          For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
          Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
          So $f_a$ is maximum at $x=a$.






          share|cite|improve this answer









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            1 Answer
            1






            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
            Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
            So $f_a$ is maximum at $x=a$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
              Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
              So $f_a$ is maximum at $x=a$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
                Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
                So $f_a$ is maximum at $x=a$.






                share|cite|improve this answer









                $endgroup$



                For $xin (0,1)$, define $$f_a(x)=alog(x)+(1-a)log(1-x)$$
                Then $$f_a^prime(x)=frac a x -frac {1-a}{1-x}=frac {a-x} {x(1-x)}$$
                So $f_a$ is maximum at $x=a$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 29 at 5:00









                Stefan LafonStefan Lafon

                3,005212




                3,005212






























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