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Arithmetic mean geometric mean inequality unclear


proving inequality?Practicing the arithmetic-geometric means inequalityArithmetic Mean and Geometric Mean Question, Guidance NeededHow prove Reversing the Arithmetic mean – Geometric mean inequality?Mean Value Theorem and Inequality.Using arithmetic mean>geometric meanNesbitt's Inequality $fracab+c+fracbc+a+fracca+bgeqfrac32$Problem in Arithmetic Mean - Geometric Mean inequalityProving Cauchy-Schwarz with Arithmetic Geometric meanInequality involving a kind of Harmonic mean













2












$begingroup$


I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ but I am wondering how the second comes from the first? could anyone explain this for me please?










      share|cite|improve this question











      $endgroup$




      I know that the AM-GM inequality takes the form $$ fracx + y2 geq sqrtxy,$$ but I read in a book another form which is $$ fracx^2 + y^22 geq |xy|,$$ 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















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 6 hours ago









      Bernard

      123k741117




      123k741117










      asked 6 hours ago









      hopefullyhopefully

      279114




      279114




















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
          $$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
          which is the second inequality (modulo capitalization).






          share|cite|improve this answer









          $endgroup$




















            5












            $begingroup$

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



            $frac1n(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



            $frac1n(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









            $endgroup$












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              2 Answers
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              2 Answers
              2






              active

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              active

              oldest

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              6












              $begingroup$

              If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
              $$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
              which is the second inequality (modulo capitalization).






              share|cite|improve this answer









              $endgroup$

















                6












                $begingroup$

                If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
                $$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
                which is the second inequality (modulo capitalization).






                share|cite|improve this answer









                $endgroup$















                  6












                  6








                  6





                  $begingroup$

                  If you plug $x=X^2$, $y=Y^2$ into the first inequality you get
                  $$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
                  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
                  $$fracX^2+Y^22 ge sqrtX^2Y^2 = sqrt(XY)^2=|XY|,$$
                  which is the second inequality (modulo capitalization).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  jgonjgon

                  16k32143




                  16k32143





















                      5












                      $begingroup$

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



                      $frac1n(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



                      $frac1n(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









                      $endgroup$

















                        5












                        $begingroup$

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



                        $frac1n(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



                        $frac1n(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









                        $endgroup$















                          5












                          5








                          5





                          $begingroup$

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



                          $frac1n(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



                          $frac1n(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



                          $frac1n(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



                          $frac1n(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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