Prove an inequality concerning $ln$ function












3












$begingroup$


For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
$$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$



I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.






real-analysis derivatives inequality







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$endgroup$

















    3












    $begingroup$


    For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
    $$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$



    I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.






    real-analysis derivatives inequality







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
      $$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$



      I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.






      real-analysis derivatives inequality







      share|cite|improve this question











      $endgroup$




      For any real number $p in (0,1)$, $delta in (0,1)$ and $p+delta <1$, prove the following inequality:
      $$(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p} ge 2delta^2$$



      I have tried some appoximation on $ln$ function such as $ln(1+x) le x$ and $ln(1+x)ge x-x^2/2$ but it doesn't work.






      real-analysis derivatives inequality




      real-analysis derivatives inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 5:55









      Michael Rozenberg

      108k1895200




      108k1895200










      asked Jan 6 at 3:51









      zbh2047zbh2047

      568




      568






















          1 Answer
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          2












          $begingroup$

          The hint.



          We need to prove that $f(delta)geq0,$ where
          $$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
          But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
          $$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
          $$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
          Can you end it now?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh! Using Taylor expansion conclude the proof. Thanks!
            $endgroup$
            – zbh2047
            Jan 6 at 8:05










          • $begingroup$
            @zbh2047 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Jan 6 at 8:55











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          1 Answer
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          active

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          active

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          active

          oldest

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          2












          $begingroup$

          The hint.



          We need to prove that $f(delta)geq0,$ where
          $$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
          But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
          $$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
          $$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
          Can you end it now?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh! Using Taylor expansion conclude the proof. Thanks!
            $endgroup$
            – zbh2047
            Jan 6 at 8:05










          • $begingroup$
            @zbh2047 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Jan 6 at 8:55
















          2












          $begingroup$

          The hint.



          We need to prove that $f(delta)geq0,$ where
          $$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
          But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
          $$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
          $$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
          Can you end it now?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh! Using Taylor expansion conclude the proof. Thanks!
            $endgroup$
            – zbh2047
            Jan 6 at 8:05










          • $begingroup$
            @zbh2047 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Jan 6 at 8:55














          2












          2








          2





          $begingroup$

          The hint.



          We need to prove that $f(delta)geq0,$ where
          $$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
          But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
          $$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
          $$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
          Can you end it now?






          share|cite|improve this answer











          $endgroup$



          The hint.



          We need to prove that $f(delta)geq0,$ where
          $$f(delta)=(p+delta)ln frac {p+delta}{p} +(1-p-delta)ln frac {1-p-delta}{1-p}- 2delta^2.$$
          But by C-S $$f''(delta)=left(lnfrac{p+delta}{p}-lnfrac{1-p-delta}{1-p}-4deltaright)'=$$
          $$=frac{1}{p+delta}+frac{1}{1-p-delta}-4geqfrac{(1+1)^2}{p+delta+1-p-delta}-4=0$$ and
          $$limlimits_{deltarightarrow0^+} f'(delta)=limlimits_{deltarightarrow0^+} f(delta)=0.$$
          Can you end it now?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 6:15

























          answered Jan 6 at 5:53









          Michael RozenbergMichael Rozenberg

          108k1895200




          108k1895200












          • $begingroup$
            Oh! Using Taylor expansion conclude the proof. Thanks!
            $endgroup$
            – zbh2047
            Jan 6 at 8:05










          • $begingroup$
            @zbh2047 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Jan 6 at 8:55


















          • $begingroup$
            Oh! Using Taylor expansion conclude the proof. Thanks!
            $endgroup$
            – zbh2047
            Jan 6 at 8:05










          • $begingroup$
            @zbh2047 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Jan 6 at 8:55
















          $begingroup$
          Oh! Using Taylor expansion conclude the proof. Thanks!
          $endgroup$
          – zbh2047
          Jan 6 at 8:05




          $begingroup$
          Oh! Using Taylor expansion conclude the proof. Thanks!
          $endgroup$
          – zbh2047
          Jan 6 at 8:05












          $begingroup$
          @zbh2047 You are welcome!
          $endgroup$
          – Michael Rozenberg
          Jan 6 at 8:55




          $begingroup$
          @zbh2047 You are welcome!
          $endgroup$
          – Michael Rozenberg
          Jan 6 at 8:55


















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