Arithmetic mean geometric mean inequality unclear












1












$begingroup$


I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?






calculus inequality







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    1












    $begingroup$


    I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?






    calculus inequality







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?






      calculus inequality







      share|cite|improve this question











      $endgroup$




      I know that the AM-GM inequality takes the form $$ frac{x + y}{2} geq sqrt{xy},$$ but I read in a book another form which is $$ frac{x^2 + y^2}{2} geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?






      calculus inequality




      calculus inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 5 hours ago









      Bernard

      123k741117




      123k741117










      asked 5 hours ago









      hopefullyhopefully

      274114




      274114






















          2 Answers
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          $begingroup$

          If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
          $$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
          which is the second inequality (modulo capitalization).






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            The AM-GM inequality for $n$ non-negative values is



            $frac1{n}(sum_{k=1}^n x_k)
            ge (prod_{k=1}^n x_k)^{1/n}
            $            .



            This can be rewritten in two ways.



            First,
            by simple algebra,



            $(sum_{k=1}^n x_i)^n
            ge n^n(prod_{k=1}^n x_k)
            $            .



            Second,
            letting $x_k = y_k^n$,
            this becomes



            $frac1{n}(sum_{k=1}^n y_k^n)
            ge prod_{k=1}^n y_k
            $            .



            It is useful to recognize
            these disguises.






            share|cite|improve this answer









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              5












              $begingroup$

              If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
              $$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
              which is the second inequality (modulo capitalization).






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
                $$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
                which is the second inequality (modulo capitalization).






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
                  $$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
                  which is the second inequality (modulo capitalization).






                  share|cite|improve this answer









                  $endgroup$



                  If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
                  $$frac{X^2+Y^2}{2} ge sqrt{X^2Y^2} = sqrt{(XY)^2}=|XY|,$$
                  which is the second inequality (modulo capitalization).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 5 hours ago









                  jgonjgon

                  16k32143




                  16k32143























                      4












                      $begingroup$

                      The AM-GM inequality for $n$ non-negative values is



                      $frac1{n}(sum_{k=1}^n x_k)
                      ge (prod_{k=1}^n x_k)^{1/n}
                      $                      .



                      This can be rewritten in two ways.



                      First,
                      by simple algebra,



                      $(sum_{k=1}^n x_i)^n
                      ge n^n(prod_{k=1}^n x_k)
                      $                      .



                      Second,
                      letting $x_k = y_k^n$,
                      this becomes



                      $frac1{n}(sum_{k=1}^n y_k^n)
                      ge prod_{k=1}^n y_k
                      $                      .



                      It is useful to recognize
                      these disguises.






                      share|cite|improve this answer









                      $endgroup$


















                        4












                        $begingroup$

                        The AM-GM inequality for $n$ non-negative values is



                        $frac1{n}(sum_{k=1}^n x_k)
                        ge (prod_{k=1}^n x_k)^{1/n}
                        $                        .



                        This can be rewritten in two ways.



                        First,
                        by simple algebra,



                        $(sum_{k=1}^n x_i)^n
                        ge n^n(prod_{k=1}^n x_k)
                        $                        .



                        Second,
                        letting $x_k = y_k^n$,
                        this becomes



                        $frac1{n}(sum_{k=1}^n y_k^n)
                        ge prod_{k=1}^n y_k
                        $                        .



                        It is useful to recognize
                        these disguises.






                        share|cite|improve this answer









                        $endgroup$
















                          4












                          4








                          4





                          $begingroup$

                          The AM-GM inequality for $n$ non-negative values is



                          $frac1{n}(sum_{k=1}^n x_k)
                          ge (prod_{k=1}^n x_k)^{1/n}
                          $                          .



                          This can be rewritten in two ways.



                          First,
                          by simple algebra,



                          $(sum_{k=1}^n x_i)^n
                          ge n^n(prod_{k=1}^n x_k)
                          $                          .



                          Second,
                          letting $x_k = y_k^n$,
                          this becomes



                          $frac1{n}(sum_{k=1}^n y_k^n)
                          ge prod_{k=1}^n y_k
                          $                          .



                          It is useful to recognize
                          these disguises.






                          share|cite|improve this answer









                          $endgroup$



                          The AM-GM inequality for $n$ non-negative values is



                          $frac1{n}(sum_{k=1}^n x_k)
                          ge (prod_{k=1}^n x_k)^{1/n}
                          $                          .



                          This can be rewritten in two ways.



                          First,
                          by simple algebra,



                          $(sum_{k=1}^n x_i)^n
                          ge n^n(prod_{k=1}^n x_k)
                          $                          .



                          Second,
                          letting $x_k = y_k^n$,
                          this becomes



                          $frac1{n}(sum_{k=1}^n y_k^n)
                          ge prod_{k=1}^n y_k
                          $                          .



                          It is useful to recognize
                          these disguises.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 5 hours ago









                          marty cohenmarty cohen

                          74.9k549130




                          74.9k549130






























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