minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$












1












$begingroup$



If $a,b,cinmathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is




what i try



Cauchy schwarz inequality



$$(a^2+(2b)^2+c^2)(2^2+frac{1}{2^2}+3^2)geq (2a+b+3c)^2$$



How do i solve it without Cauchy schwarz inequality Help me please






inequality optimization quadratics maxima-minima







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    1












    $begingroup$



    If $a,b,cinmathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is




    what i try



    Cauchy schwarz inequality



    $$(a^2+(2b)^2+c^2)(2^2+frac{1}{2^2}+3^2)geq (2a+b+3c)^2$$



    How do i solve it without Cauchy schwarz inequality Help me please






    inequality optimization quadratics maxima-minima







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      If $a,b,cinmathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is




      what i try



      Cauchy schwarz inequality



      $$(a^2+(2b)^2+c^2)(2^2+frac{1}{2^2}+3^2)geq (2a+b+3c)^2$$



      How do i solve it without Cauchy schwarz inequality Help me please






      inequality optimization quadratics maxima-minima







      share|cite|improve this question











      $endgroup$





      If $a,b,cinmathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is




      what i try



      Cauchy schwarz inequality



      $$(a^2+(2b)^2+c^2)(2^2+frac{1}{2^2}+3^2)geq (2a+b+3c)^2$$



      How do i solve it without Cauchy schwarz inequality Help me please






      inequality optimization quadratics maxima-minima




      inequality optimization quadratics maxima-minima






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      share|cite|improve this question













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      edited 2 days ago









      user21820

      39.8k544158




      39.8k544158










      asked 2 days ago









      jackyjacky

      1,341816




      1,341816






















          4 Answers
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          $begingroup$

          Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)in P$ for $tinBbb R$, then we get $53t=40$ and $x=frac{40}{53}(4,1,6)$ follows. Hence we obtain
          $$
          a^2+(b')^2+c^2 ge |x|^2=frac{1600}{53}
          $$           is the minimum.






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            An alternative way of solving the problem is using Lagrange multipliers, other than that (i.e. without using the CS inequality), I don't see a simple solution, and neither do I see a reason to look for such a solution.






            share|cite|improve this answer









            $endgroup$





















              4












              $begingroup$

              I think this solution is more simple.



              From $$2a+b+3c=20Leftrightarrow b=20-2a-3c$$



              Then $$A=a^2+4left(20-2a-3cright)^2+c^2$$



              $$=17a^2+48ac-320a+37c^2-480c+1600$$



              $$=frac{1}{17}(17a+24c-160)^2+frac{1}{17}left(53left(c-frac{240}{53}right)^2+frac{27200}{53}right )$$



              $$ge frac{1}{17}cdot frac{27200}{53}=frac{1600}{53}$$



              The equality occurs when $(a;b;c)=left(frac{160}{53};frac{20}{53};frac{240}{53}right)$






              share|cite|improve this answer









              $endgroup$





















                3












                $begingroup$

                Let $(a,b,c)=left(frac{160}{53},frac{20}{53},frac{240}{53}right).$



                Thus, we get a value $frac{1600}{53}.$



                We'll prove that it's a minimal value.



                Indeed, we need to prove that
                $$a^2+4b^2+c^2geqfrac{1600}{53}cdotleft(frac{2a+b+3c}{20}right)^2$$ or
                $$37a^2-16(b+3c)a+208b^2-24bc+17c^2geq0,$$ for which it's enough to prove that
                $$64(b+3c)^2-37(208b^2-24bc+17c^2)leq0,$$ which is
                $$(12b-c)^2geq0.$$
                Done!






                share|cite|improve this answer









                $endgroup$














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                  4 Answers
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                  active

                  oldest

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                  4 Answers
                  4






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                  oldest

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                  active

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                  active

                  oldest

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                  1












                  $begingroup$

                  Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)in P$ for $tinBbb R$, then we get $53t=40$ and $x=frac{40}{53}(4,1,6)$ follows. Hence we obtain
                  $$
                  a^2+(b')^2+c^2 ge |x|^2=frac{1600}{53}
                  $$                   is the minimum.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)in P$ for $tinBbb R$, then we get $53t=40$ and $x=frac{40}{53}(4,1,6)$ follows. Hence we obtain
                    $$
                    a^2+(b')^2+c^2 ge |x|^2=frac{1600}{53}
                    $$                     is the minimum.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)in P$ for $tinBbb R$, then we get $53t=40$ and $x=frac{40}{53}(4,1,6)$ follows. Hence we obtain
                      $$
                      a^2+(b')^2+c^2 ge |x|^2=frac{1600}{53}
                      $$                       is the minimum.






                      share|cite|improve this answer









                      $endgroup$



                      Put $b'=2b$ so that we minimize $a^2+(b')^2+c^2$ subject to $4a+b'+6c=40$. Now, since the point $x=(a,b',c)$ is on the plane $P: 4x+y+6z=40$, its distance from the origin is minimized when it is parallel to the normal vector $n=(4,1,6)$ to the plane $P$. If we solve $x=tn=(4t,t,6t)in P$ for $tinBbb R$, then we get $53t=40$ and $x=frac{40}{53}(4,1,6)$ follows. Hence we obtain
                      $$
                      a^2+(b')^2+c^2 ge |x|^2=frac{1600}{53}
                      $$                       is the minimum.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 days ago









                      SongSong

                      18.5k21651




                      18.5k21651























                          4












                          $begingroup$

                          An alternative way of solving the problem is using Lagrange multipliers, other than that (i.e. without using the CS inequality), I don't see a simple solution, and neither do I see a reason to look for such a solution.






                          share|cite|improve this answer









                          $endgroup$


















                            4












                            $begingroup$

                            An alternative way of solving the problem is using Lagrange multipliers, other than that (i.e. without using the CS inequality), I don't see a simple solution, and neither do I see a reason to look for such a solution.






                            share|cite|improve this answer









                            $endgroup$
















                              4












                              4








                              4





                              $begingroup$

                              An alternative way of solving the problem is using Lagrange multipliers, other than that (i.e. without using the CS inequality), I don't see a simple solution, and neither do I see a reason to look for such a solution.






                              share|cite|improve this answer









                              $endgroup$



                              An alternative way of solving the problem is using Lagrange multipliers, other than that (i.e. without using the CS inequality), I don't see a simple solution, and neither do I see a reason to look for such a solution.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 days ago









                              5xum5xum

                              91.8k394161




                              91.8k394161























                                  4












                                  $begingroup$

                                  I think this solution is more simple.



                                  From $$2a+b+3c=20Leftrightarrow b=20-2a-3c$$



                                  Then $$A=a^2+4left(20-2a-3cright)^2+c^2$$



                                  $$=17a^2+48ac-320a+37c^2-480c+1600$$



                                  $$=frac{1}{17}(17a+24c-160)^2+frac{1}{17}left(53left(c-frac{240}{53}right)^2+frac{27200}{53}right )$$



                                  $$ge frac{1}{17}cdot frac{27200}{53}=frac{1600}{53}$$



                                  The equality occurs when $(a;b;c)=left(frac{160}{53};frac{20}{53};frac{240}{53}right)$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    4












                                    $begingroup$

                                    I think this solution is more simple.



                                    From $$2a+b+3c=20Leftrightarrow b=20-2a-3c$$



                                    Then $$A=a^2+4left(20-2a-3cright)^2+c^2$$



                                    $$=17a^2+48ac-320a+37c^2-480c+1600$$



                                    $$=frac{1}{17}(17a+24c-160)^2+frac{1}{17}left(53left(c-frac{240}{53}right)^2+frac{27200}{53}right )$$



                                    $$ge frac{1}{17}cdot frac{27200}{53}=frac{1600}{53}$$



                                    The equality occurs when $(a;b;c)=left(frac{160}{53};frac{20}{53};frac{240}{53}right)$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      4












                                      4








                                      4





                                      $begingroup$

                                      I think this solution is more simple.



                                      From $$2a+b+3c=20Leftrightarrow b=20-2a-3c$$



                                      Then $$A=a^2+4left(20-2a-3cright)^2+c^2$$



                                      $$=17a^2+48ac-320a+37c^2-480c+1600$$



                                      $$=frac{1}{17}(17a+24c-160)^2+frac{1}{17}left(53left(c-frac{240}{53}right)^2+frac{27200}{53}right )$$



                                      $$ge frac{1}{17}cdot frac{27200}{53}=frac{1600}{53}$$



                                      The equality occurs when $(a;b;c)=left(frac{160}{53};frac{20}{53};frac{240}{53}right)$






                                      share|cite|improve this answer









                                      $endgroup$



                                      I think this solution is more simple.



                                      From $$2a+b+3c=20Leftrightarrow b=20-2a-3c$$



                                      Then $$A=a^2+4left(20-2a-3cright)^2+c^2$$



                                      $$=17a^2+48ac-320a+37c^2-480c+1600$$



                                      $$=frac{1}{17}(17a+24c-160)^2+frac{1}{17}left(53left(c-frac{240}{53}right)^2+frac{27200}{53}right )$$



                                      $$ge frac{1}{17}cdot frac{27200}{53}=frac{1600}{53}$$



                                      The equality occurs when $(a;b;c)=left(frac{160}{53};frac{20}{53};frac{240}{53}right)$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 2 days ago









                                      Word ShallowWord Shallow

                                      1,1182621




                                      1,1182621























                                          3












                                          $begingroup$

                                          Let $(a,b,c)=left(frac{160}{53},frac{20}{53},frac{240}{53}right).$



                                          Thus, we get a value $frac{1600}{53}.$



                                          We'll prove that it's a minimal value.



                                          Indeed, we need to prove that
                                          $$a^2+4b^2+c^2geqfrac{1600}{53}cdotleft(frac{2a+b+3c}{20}right)^2$$ or
                                          $$37a^2-16(b+3c)a+208b^2-24bc+17c^2geq0,$$ for which it's enough to prove that
                                          $$64(b+3c)^2-37(208b^2-24bc+17c^2)leq0,$$ which is
                                          $$(12b-c)^2geq0.$$
                                          Done!






                                          share|cite|improve this answer









                                          $endgroup$


















                                            3












                                            $begingroup$

                                            Let $(a,b,c)=left(frac{160}{53},frac{20}{53},frac{240}{53}right).$



                                            Thus, we get a value $frac{1600}{53}.$



                                            We'll prove that it's a minimal value.



                                            Indeed, we need to prove that
                                            $$a^2+4b^2+c^2geqfrac{1600}{53}cdotleft(frac{2a+b+3c}{20}right)^2$$ or
                                            $$37a^2-16(b+3c)a+208b^2-24bc+17c^2geq0,$$ for which it's enough to prove that
                                            $$64(b+3c)^2-37(208b^2-24bc+17c^2)leq0,$$ which is
                                            $$(12b-c)^2geq0.$$
                                            Done!






                                            share|cite|improve this answer









                                            $endgroup$
















                                              3












                                              3








                                              3





                                              $begingroup$

                                              Let $(a,b,c)=left(frac{160}{53},frac{20}{53},frac{240}{53}right).$



                                              Thus, we get a value $frac{1600}{53}.$



                                              We'll prove that it's a minimal value.



                                              Indeed, we need to prove that
                                              $$a^2+4b^2+c^2geqfrac{1600}{53}cdotleft(frac{2a+b+3c}{20}right)^2$$ or
                                              $$37a^2-16(b+3c)a+208b^2-24bc+17c^2geq0,$$ for which it's enough to prove that
                                              $$64(b+3c)^2-37(208b^2-24bc+17c^2)leq0,$$ which is
                                              $$(12b-c)^2geq0.$$
                                              Done!






                                              share|cite|improve this answer









                                              $endgroup$



                                              Let $(a,b,c)=left(frac{160}{53},frac{20}{53},frac{240}{53}right).$



                                              Thus, we get a value $frac{1600}{53}.$



                                              We'll prove that it's a minimal value.



                                              Indeed, we need to prove that
                                              $$a^2+4b^2+c^2geqfrac{1600}{53}cdotleft(frac{2a+b+3c}{20}right)^2$$ or
                                              $$37a^2-16(b+3c)a+208b^2-24bc+17c^2geq0,$$ for which it's enough to prove that
                                              $$64(b+3c)^2-37(208b^2-24bc+17c^2)leq0,$$ which is
                                              $$(12b-c)^2geq0.$$
                                              Done!







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 2 days ago









                                              Michael RozenbergMichael Rozenberg

                                              109k1896201




                                              109k1896201






























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