Using AM-GM inequality to prove












1












$begingroup$


prove that $$x^4 + y^4 + z^4 geq xyz(x+y+z)$$



This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.






proof-explanation







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$endgroup$

















    1












    $begingroup$


    prove that $$x^4 + y^4 + z^4 geq xyz(x+y+z)$$



    This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.






    proof-explanation







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      prove that $$x^4 + y^4 + z^4 geq xyz(x+y+z)$$



      This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.






      proof-explanation







      share|cite|improve this question











      $endgroup$




      prove that $$x^4 + y^4 + z^4 geq xyz(x+y+z)$$



      This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.






      proof-explanation




      proof-explanation






      share|cite|improve this question















      share|cite|improve this question













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      edited Jun 9 '16 at 17:36









      danimal

      1,8161612




      1,8161612










      asked Jun 9 '16 at 17:32









      Gigi ChuckGigi Chuck

      828




      828






















          3 Answers
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          $begingroup$

          Use: $$a^2+b^2+c^2ge ab+bc+ca$$
          Proof:$$2(a^2+b^2+c^2)ge 2(ab+bc+ca) Leftrightarrow$$
          $$Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2ge0$$



          Then
          $$x^4+y^4+z^4ge x^2y^2+y^2z^2+z^2x^2ge xy^2z+x^2yz+xyz^2=$$
          $$=xyz(x+y+z)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how did the x^2 change to xz?
            $endgroup$
            – Gigi Chuck
            Jun 9 '16 at 17:41






          • 1




            $begingroup$
            we use double $a^2+b^2+c^2ge ab+bc+ca$ Then $$x^2y^2+y^2z^2+z^2x^2 ge xy^2z+yz^2x+x^2yz$$
            $endgroup$
            – Roman83
            Jun 9 '16 at 17:43





















          1












          $begingroup$

          AM–GM is invoked in two steps as follows:



          $begin{array}{rcl}x^4+y^4+z^4 &=& dfrac{x^4+y^4}2+dfrac{y^4+z^4}2+dfrac{z^4+x^4}2 \\ &ge& x^2y^2+y^2z^2+z^2x^2 \\ &=& dfrac{x^2y^2+y^2z^2}2+dfrac{y^2z^2+z^2x^2}2+dfrac{z^2x^2+x^2y^2}2 \\ &ge& xy^2z+yz^2x+zx^2y \\ &=& xyz(x+y+z)end{array}$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            The inequality is equivalent to ( by expanding RHS)
            $$ x^4 + y^4 +z^4 ge x^2yz + y^2xz + z^2xy$$
            By AM-GM



            $$ x^4+x^4 +y^4 +z^4 ge 4(x^8y^4z^4)^frac{1}{4} = 4x^2yz$$
            $$ y^4+y^4 +x^4 +z^4 ge 4(y^8x^4z^4)^frac{1}{4} = 4y^2xz$$
            $$ z^4+z^4 +x^4 +y^4 ge 4(z^8x^4y^4)^frac{1}{4} = 4z^2xy$$
            Adding them up we get



            $$4(x^4 +y^4 +z^4) ge 4(x^2yz+y^2xz + z^2xy)$$
            $$ x^4 +y^4 +z^4 ge x^2yz+y^2xz + z^2xy $$






            share|cite|improve this answer









            $endgroup$













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              3 Answers
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              3 Answers
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              3












              $begingroup$

              Use: $$a^2+b^2+c^2ge ab+bc+ca$$
              Proof:$$2(a^2+b^2+c^2)ge 2(ab+bc+ca) Leftrightarrow$$
              $$Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2ge0$$



              Then
              $$x^4+y^4+z^4ge x^2y^2+y^2z^2+z^2x^2ge xy^2z+x^2yz+xyz^2=$$
              $$=xyz(x+y+z)$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                how did the x^2 change to xz?
                $endgroup$
                – Gigi Chuck
                Jun 9 '16 at 17:41






              • 1




                $begingroup$
                we use double $a^2+b^2+c^2ge ab+bc+ca$ Then $$x^2y^2+y^2z^2+z^2x^2 ge xy^2z+yz^2x+x^2yz$$
                $endgroup$
                – Roman83
                Jun 9 '16 at 17:43


















              3












              $begingroup$

              Use: $$a^2+b^2+c^2ge ab+bc+ca$$
              Proof:$$2(a^2+b^2+c^2)ge 2(ab+bc+ca) Leftrightarrow$$
              $$Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2ge0$$



              Then
              $$x^4+y^4+z^4ge x^2y^2+y^2z^2+z^2x^2ge xy^2z+x^2yz+xyz^2=$$
              $$=xyz(x+y+z)$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                how did the x^2 change to xz?
                $endgroup$
                – Gigi Chuck
                Jun 9 '16 at 17:41






              • 1




                $begingroup$
                we use double $a^2+b^2+c^2ge ab+bc+ca$ Then $$x^2y^2+y^2z^2+z^2x^2 ge xy^2z+yz^2x+x^2yz$$
                $endgroup$
                – Roman83
                Jun 9 '16 at 17:43
















              3












              3








              3





              $begingroup$

              Use: $$a^2+b^2+c^2ge ab+bc+ca$$
              Proof:$$2(a^2+b^2+c^2)ge 2(ab+bc+ca) Leftrightarrow$$
              $$Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2ge0$$



              Then
              $$x^4+y^4+z^4ge x^2y^2+y^2z^2+z^2x^2ge xy^2z+x^2yz+xyz^2=$$
              $$=xyz(x+y+z)$$






              share|cite|improve this answer









              $endgroup$



              Use: $$a^2+b^2+c^2ge ab+bc+ca$$
              Proof:$$2(a^2+b^2+c^2)ge 2(ab+bc+ca) Leftrightarrow$$
              $$Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2ge0$$



              Then
              $$x^4+y^4+z^4ge x^2y^2+y^2z^2+z^2x^2ge xy^2z+x^2yz+xyz^2=$$
              $$=xyz(x+y+z)$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jun 9 '16 at 17:36









              Roman83Roman83

              14.4k31956




              14.4k31956












              • $begingroup$
                how did the x^2 change to xz?
                $endgroup$
                – Gigi Chuck
                Jun 9 '16 at 17:41






              • 1




                $begingroup$
                we use double $a^2+b^2+c^2ge ab+bc+ca$ Then $$x^2y^2+y^2z^2+z^2x^2 ge xy^2z+yz^2x+x^2yz$$
                $endgroup$
                – Roman83
                Jun 9 '16 at 17:43




















              • $begingroup$
                how did the x^2 change to xz?
                $endgroup$
                – Gigi Chuck
                Jun 9 '16 at 17:41






              • 1




                $begingroup$
                we use double $a^2+b^2+c^2ge ab+bc+ca$ Then $$x^2y^2+y^2z^2+z^2x^2 ge xy^2z+yz^2x+x^2yz$$
                $endgroup$
                – Roman83
                Jun 9 '16 at 17:43


















              $begingroup$
              how did the x^2 change to xz?
              $endgroup$
              – Gigi Chuck
              Jun 9 '16 at 17:41




              $begingroup$
              how did the x^2 change to xz?
              $endgroup$
              – Gigi Chuck
              Jun 9 '16 at 17:41




              1




              1




              $begingroup$
              we use double $a^2+b^2+c^2ge ab+bc+ca$ Then $$x^2y^2+y^2z^2+z^2x^2 ge xy^2z+yz^2x+x^2yz$$
              $endgroup$
              – Roman83
              Jun 9 '16 at 17:43






              $begingroup$
              we use double $a^2+b^2+c^2ge ab+bc+ca$ Then $$x^2y^2+y^2z^2+z^2x^2 ge xy^2z+yz^2x+x^2yz$$
              $endgroup$
              – Roman83
              Jun 9 '16 at 17:43













              1












              $begingroup$

              AM–GM is invoked in two steps as follows:



              $begin{array}{rcl}x^4+y^4+z^4 &=& dfrac{x^4+y^4}2+dfrac{y^4+z^4}2+dfrac{z^4+x^4}2 \\ &ge& x^2y^2+y^2z^2+z^2x^2 \\ &=& dfrac{x^2y^2+y^2z^2}2+dfrac{y^2z^2+z^2x^2}2+dfrac{z^2x^2+x^2y^2}2 \\ &ge& xy^2z+yz^2x+zx^2y \\ &=& xyz(x+y+z)end{array}$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                AM–GM is invoked in two steps as follows:



                $begin{array}{rcl}x^4+y^4+z^4 &=& dfrac{x^4+y^4}2+dfrac{y^4+z^4}2+dfrac{z^4+x^4}2 \\ &ge& x^2y^2+y^2z^2+z^2x^2 \\ &=& dfrac{x^2y^2+y^2z^2}2+dfrac{y^2z^2+z^2x^2}2+dfrac{z^2x^2+x^2y^2}2 \\ &ge& xy^2z+yz^2x+zx^2y \\ &=& xyz(x+y+z)end{array}$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  AM–GM is invoked in two steps as follows:



                  $begin{array}{rcl}x^4+y^4+z^4 &=& dfrac{x^4+y^4}2+dfrac{y^4+z^4}2+dfrac{z^4+x^4}2 \\ &ge& x^2y^2+y^2z^2+z^2x^2 \\ &=& dfrac{x^2y^2+y^2z^2}2+dfrac{y^2z^2+z^2x^2}2+dfrac{z^2x^2+x^2y^2}2 \\ &ge& xy^2z+yz^2x+zx^2y \\ &=& xyz(x+y+z)end{array}$






                  share|cite|improve this answer









                  $endgroup$



                  AM–GM is invoked in two steps as follows:



                  $begin{array}{rcl}x^4+y^4+z^4 &=& dfrac{x^4+y^4}2+dfrac{y^4+z^4}2+dfrac{z^4+x^4}2 \\ &ge& x^2y^2+y^2z^2+z^2x^2 \\ &=& dfrac{x^2y^2+y^2z^2}2+dfrac{y^2z^2+z^2x^2}2+dfrac{z^2x^2+x^2y^2}2 \\ &ge& xy^2z+yz^2x+zx^2y \\ &=& xyz(x+y+z)end{array}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jun 9 '16 at 17:57









                  George LawGeorge Law

                  3,58711423




                  3,58711423























                      0












                      $begingroup$

                      The inequality is equivalent to ( by expanding RHS)
                      $$ x^4 + y^4 +z^4 ge x^2yz + y^2xz + z^2xy$$
                      By AM-GM



                      $$ x^4+x^4 +y^4 +z^4 ge 4(x^8y^4z^4)^frac{1}{4} = 4x^2yz$$
                      $$ y^4+y^4 +x^4 +z^4 ge 4(y^8x^4z^4)^frac{1}{4} = 4y^2xz$$
                      $$ z^4+z^4 +x^4 +y^4 ge 4(z^8x^4y^4)^frac{1}{4} = 4z^2xy$$
                      Adding them up we get



                      $$4(x^4 +y^4 +z^4) ge 4(x^2yz+y^2xz + z^2xy)$$
                      $$ x^4 +y^4 +z^4 ge x^2yz+y^2xz + z^2xy $$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        The inequality is equivalent to ( by expanding RHS)
                        $$ x^4 + y^4 +z^4 ge x^2yz + y^2xz + z^2xy$$
                        By AM-GM



                        $$ x^4+x^4 +y^4 +z^4 ge 4(x^8y^4z^4)^frac{1}{4} = 4x^2yz$$
                        $$ y^4+y^4 +x^4 +z^4 ge 4(y^8x^4z^4)^frac{1}{4} = 4y^2xz$$
                        $$ z^4+z^4 +x^4 +y^4 ge 4(z^8x^4y^4)^frac{1}{4} = 4z^2xy$$
                        Adding them up we get



                        $$4(x^4 +y^4 +z^4) ge 4(x^2yz+y^2xz + z^2xy)$$
                        $$ x^4 +y^4 +z^4 ge x^2yz+y^2xz + z^2xy $$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The inequality is equivalent to ( by expanding RHS)
                          $$ x^4 + y^4 +z^4 ge x^2yz + y^2xz + z^2xy$$
                          By AM-GM



                          $$ x^4+x^4 +y^4 +z^4 ge 4(x^8y^4z^4)^frac{1}{4} = 4x^2yz$$
                          $$ y^4+y^4 +x^4 +z^4 ge 4(y^8x^4z^4)^frac{1}{4} = 4y^2xz$$
                          $$ z^4+z^4 +x^4 +y^4 ge 4(z^8x^4y^4)^frac{1}{4} = 4z^2xy$$
                          Adding them up we get



                          $$4(x^4 +y^4 +z^4) ge 4(x^2yz+y^2xz + z^2xy)$$
                          $$ x^4 +y^4 +z^4 ge x^2yz+y^2xz + z^2xy $$






                          share|cite|improve this answer









                          $endgroup$



                          The inequality is equivalent to ( by expanding RHS)
                          $$ x^4 + y^4 +z^4 ge x^2yz + y^2xz + z^2xy$$
                          By AM-GM



                          $$ x^4+x^4 +y^4 +z^4 ge 4(x^8y^4z^4)^frac{1}{4} = 4x^2yz$$
                          $$ y^4+y^4 +x^4 +z^4 ge 4(y^8x^4z^4)^frac{1}{4} = 4y^2xz$$
                          $$ z^4+z^4 +x^4 +y^4 ge 4(z^8x^4y^4)^frac{1}{4} = 4z^2xy$$
                          Adding them up we get



                          $$4(x^4 +y^4 +z^4) ge 4(x^2yz+y^2xz + z^2xy)$$
                          $$ x^4 +y^4 +z^4 ge x^2yz+y^2xz + z^2xy $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 22 at 0:34









                          Ehit KarimEhit Karim

                          341




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