Is it based on Tchebyshev?











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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?



I tried with Tchebyshev inequality on sets ${a, b, c}$ and ${a^2, b^2 , c^2}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.



I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.










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    up vote
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    down vote

    favorite












    If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?



    I tried with Tchebyshev inequality on sets ${a, b, c}$ and ${a^2, b^2 , c^2}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.



    I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.










    share|cite|improve this question


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?



      I tried with Tchebyshev inequality on sets ${a, b, c}$ and ${a^2, b^2 , c^2}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.



      I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.










      share|cite|improve this question















      If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?



      I tried with Tchebyshev inequality on sets ${a, b, c}$ and ${a^2, b^2 , c^2}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.



      I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.







      inequality






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      edited Nov 16 at 13:05









      amWhy

      191k27223438




      191k27223438










      asked Nov 16 at 11:47









      Ramanujam Ganit Prashikshan Ke

      1267




      1267






















          4 Answers
          4






          active

          oldest

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          up vote
          1
          down vote



          accepted










          You can do it using the mean of the $m$ power for $m=frac{3}{2}$ (or convexity of $x^frac{3}{2}$ for that matter) :



          We have that $frac{x^{frac{3}{2}}+y^{frac{3}{2}}+z^frac{3}{2}}{3} geq (frac{x+y+z}{3})^frac{3}{2}$ for any $x,y,z > 0$



          Substituting $x=a^2, y=b^2, z=c^2$ we obtain $frac{a^3+b^3+c^3}{3} geq (frac{a^2+b^2+c^2}{3})^frac{3}{2} = 9^frac{3}{2}=27$ and therefore the sought minimum is $3cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.






          share|cite|improve this answer






























            up vote
            2
            down vote













            From the power mean inequality we know that



            $$sqrt[q]{sum_{i=1}^nw_ix_i^q} geq sqrt[p]{sum_{i=1}^nw_ix_i^p} quad forall p <q$$



            where $ sum_{i=1}^n w_i=1 $



            In your case $p=2, q=3$ and $n=3$. Hence,



            $$left(dfrac{a^3+b^3+c^3}{3}right)^{1/3}ge left(dfrac{a^2+b^2+c^2}{3}right)^{1/2}=3$$






            share|cite|improve this answer




























              up vote
              2
              down vote













              Okay, the question requires knowledge of the AM-GM inequality, which you do have.
              You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
              The answer is $81$.
              Hope this helps.






              share|cite|improve this answer






























                up vote
                0
                down vote













                For $a=b=c=3$ we obtain a value $81$.



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



                Indeed, we need to prove that $$a^3+b^3+c^3geq81,$$ which is true because
                $$a^3+b^3+c^3-81=sum_{cyc}(a^3-27)=$$
                $$=sum_{cyc}left(a^3-27-frac{9}{2}(a^2-9)right)=frac{1}{2}sum_{cyc}(a-3)^2(2a+3)geq0.$$
                Also, Holder helps:
                $$a^3+b^3+c^3=sqrt{frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}geqsqrt{frac{(a^2+b^2+c^2)^3}{3}}=81.$$






                share|cite|improve this answer























                • From which source can I understand holder's inequality with elementary illustrations and exercises. You seem to be good at this.
                  – Ramanujam Ganit Prashikshan Ke
                  Nov 16 at 16:55










                • @Ramanujam Ganit Prashikshan Ke You can read solutions with tag holder-inequality in this forum. I think it's the best illustration. See also here: math.stackexchange.com/tags/holder-inequality/info
                  – Michael Rozenberg
                  Nov 16 at 17:29













                Your Answer





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                4 Answers
                4






                active

                oldest

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                4 Answers
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                active

                oldest

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                active

                oldest

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                active

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                up vote
                1
                down vote



                accepted










                You can do it using the mean of the $m$ power for $m=frac{3}{2}$ (or convexity of $x^frac{3}{2}$ for that matter) :



                We have that $frac{x^{frac{3}{2}}+y^{frac{3}{2}}+z^frac{3}{2}}{3} geq (frac{x+y+z}{3})^frac{3}{2}$ for any $x,y,z > 0$



                Substituting $x=a^2, y=b^2, z=c^2$ we obtain $frac{a^3+b^3+c^3}{3} geq (frac{a^2+b^2+c^2}{3})^frac{3}{2} = 9^frac{3}{2}=27$ and therefore the sought minimum is $3cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.






                share|cite|improve this answer



























                  up vote
                  1
                  down vote



                  accepted










                  You can do it using the mean of the $m$ power for $m=frac{3}{2}$ (or convexity of $x^frac{3}{2}$ for that matter) :



                  We have that $frac{x^{frac{3}{2}}+y^{frac{3}{2}}+z^frac{3}{2}}{3} geq (frac{x+y+z}{3})^frac{3}{2}$ for any $x,y,z > 0$



                  Substituting $x=a^2, y=b^2, z=c^2$ we obtain $frac{a^3+b^3+c^3}{3} geq (frac{a^2+b^2+c^2}{3})^frac{3}{2} = 9^frac{3}{2}=27$ and therefore the sought minimum is $3cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote



                    accepted







                    up vote
                    1
                    down vote



                    accepted






                    You can do it using the mean of the $m$ power for $m=frac{3}{2}$ (or convexity of $x^frac{3}{2}$ for that matter) :



                    We have that $frac{x^{frac{3}{2}}+y^{frac{3}{2}}+z^frac{3}{2}}{3} geq (frac{x+y+z}{3})^frac{3}{2}$ for any $x,y,z > 0$



                    Substituting $x=a^2, y=b^2, z=c^2$ we obtain $frac{a^3+b^3+c^3}{3} geq (frac{a^2+b^2+c^2}{3})^frac{3}{2} = 9^frac{3}{2}=27$ and therefore the sought minimum is $3cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.






                    share|cite|improve this answer














                    You can do it using the mean of the $m$ power for $m=frac{3}{2}$ (or convexity of $x^frac{3}{2}$ for that matter) :



                    We have that $frac{x^{frac{3}{2}}+y^{frac{3}{2}}+z^frac{3}{2}}{3} geq (frac{x+y+z}{3})^frac{3}{2}$ for any $x,y,z > 0$



                    Substituting $x=a^2, y=b^2, z=c^2$ we obtain $frac{a^3+b^3+c^3}{3} geq (frac{a^2+b^2+c^2}{3})^frac{3}{2} = 9^frac{3}{2}=27$ and therefore the sought minimum is $3cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 17 at 13:16









                    amWhy

                    191k27223438




                    191k27223438










                    answered Nov 16 at 13:32









                    Sorin Tirc

                    66210




                    66210






















                        up vote
                        2
                        down vote













                        From the power mean inequality we know that



                        $$sqrt[q]{sum_{i=1}^nw_ix_i^q} geq sqrt[p]{sum_{i=1}^nw_ix_i^p} quad forall p <q$$



                        where $ sum_{i=1}^n w_i=1 $



                        In your case $p=2, q=3$ and $n=3$. Hence,



                        $$left(dfrac{a^3+b^3+c^3}{3}right)^{1/3}ge left(dfrac{a^2+b^2+c^2}{3}right)^{1/2}=3$$






                        share|cite|improve this answer

























                          up vote
                          2
                          down vote













                          From the power mean inequality we know that



                          $$sqrt[q]{sum_{i=1}^nw_ix_i^q} geq sqrt[p]{sum_{i=1}^nw_ix_i^p} quad forall p <q$$



                          where $ sum_{i=1}^n w_i=1 $



                          In your case $p=2, q=3$ and $n=3$. Hence,



                          $$left(dfrac{a^3+b^3+c^3}{3}right)^{1/3}ge left(dfrac{a^2+b^2+c^2}{3}right)^{1/2}=3$$






                          share|cite|improve this answer























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            From the power mean inequality we know that



                            $$sqrt[q]{sum_{i=1}^nw_ix_i^q} geq sqrt[p]{sum_{i=1}^nw_ix_i^p} quad forall p <q$$



                            where $ sum_{i=1}^n w_i=1 $



                            In your case $p=2, q=3$ and $n=3$. Hence,



                            $$left(dfrac{a^3+b^3+c^3}{3}right)^{1/3}ge left(dfrac{a^2+b^2+c^2}{3}right)^{1/2}=3$$






                            share|cite|improve this answer












                            From the power mean inequality we know that



                            $$sqrt[q]{sum_{i=1}^nw_ix_i^q} geq sqrt[p]{sum_{i=1}^nw_ix_i^p} quad forall p <q$$



                            where $ sum_{i=1}^n w_i=1 $



                            In your case $p=2, q=3$ and $n=3$. Hence,



                            $$left(dfrac{a^3+b^3+c^3}{3}right)^{1/3}ge left(dfrac{a^2+b^2+c^2}{3}right)^{1/2}=3$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 16 at 13:39









                            callculus

                            17.6k31427




                            17.6k31427






















                                up vote
                                2
                                down vote













                                Okay, the question requires knowledge of the AM-GM inequality, which you do have.
                                You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
                                The answer is $81$.
                                Hope this helps.






                                share|cite|improve this answer



























                                  up vote
                                  2
                                  down vote













                                  Okay, the question requires knowledge of the AM-GM inequality, which you do have.
                                  You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
                                  The answer is $81$.
                                  Hope this helps.






                                  share|cite|improve this answer

























                                    up vote
                                    2
                                    down vote










                                    up vote
                                    2
                                    down vote









                                    Okay, the question requires knowledge of the AM-GM inequality, which you do have.
                                    You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
                                    The answer is $81$.
                                    Hope this helps.






                                    share|cite|improve this answer














                                    Okay, the question requires knowledge of the AM-GM inequality, which you do have.
                                    You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
                                    The answer is $81$.
                                    Hope this helps.







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 17 at 13:16

























                                    answered Nov 16 at 13:23









                                    AryanSonwatikar

                                    759




                                    759






















                                        up vote
                                        0
                                        down vote













                                        For $a=b=c=3$ we obtain a value $81$.



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



                                        Indeed, we need to prove that $$a^3+b^3+c^3geq81,$$ which is true because
                                        $$a^3+b^3+c^3-81=sum_{cyc}(a^3-27)=$$
                                        $$=sum_{cyc}left(a^3-27-frac{9}{2}(a^2-9)right)=frac{1}{2}sum_{cyc}(a-3)^2(2a+3)geq0.$$
                                        Also, Holder helps:
                                        $$a^3+b^3+c^3=sqrt{frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}geqsqrt{frac{(a^2+b^2+c^2)^3}{3}}=81.$$






                                        share|cite|improve this answer























                                        • From which source can I understand holder's inequality with elementary illustrations and exercises. You seem to be good at this.
                                          – Ramanujam Ganit Prashikshan Ke
                                          Nov 16 at 16:55










                                        • @Ramanujam Ganit Prashikshan Ke You can read solutions with tag holder-inequality in this forum. I think it's the best illustration. See also here: math.stackexchange.com/tags/holder-inequality/info
                                          – Michael Rozenberg
                                          Nov 16 at 17:29

















                                        up vote
                                        0
                                        down vote













                                        For $a=b=c=3$ we obtain a value $81$.



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



                                        Indeed, we need to prove that $$a^3+b^3+c^3geq81,$$ which is true because
                                        $$a^3+b^3+c^3-81=sum_{cyc}(a^3-27)=$$
                                        $$=sum_{cyc}left(a^3-27-frac{9}{2}(a^2-9)right)=frac{1}{2}sum_{cyc}(a-3)^2(2a+3)geq0.$$
                                        Also, Holder helps:
                                        $$a^3+b^3+c^3=sqrt{frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}geqsqrt{frac{(a^2+b^2+c^2)^3}{3}}=81.$$






                                        share|cite|improve this answer























                                        • From which source can I understand holder's inequality with elementary illustrations and exercises. You seem to be good at this.
                                          – Ramanujam Ganit Prashikshan Ke
                                          Nov 16 at 16:55










                                        • @Ramanujam Ganit Prashikshan Ke You can read solutions with tag holder-inequality in this forum. I think it's the best illustration. See also here: math.stackexchange.com/tags/holder-inequality/info
                                          – Michael Rozenberg
                                          Nov 16 at 17:29















                                        up vote
                                        0
                                        down vote










                                        up vote
                                        0
                                        down vote









                                        For $a=b=c=3$ we obtain a value $81$.



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



                                        Indeed, we need to prove that $$a^3+b^3+c^3geq81,$$ which is true because
                                        $$a^3+b^3+c^3-81=sum_{cyc}(a^3-27)=$$
                                        $$=sum_{cyc}left(a^3-27-frac{9}{2}(a^2-9)right)=frac{1}{2}sum_{cyc}(a-3)^2(2a+3)geq0.$$
                                        Also, Holder helps:
                                        $$a^3+b^3+c^3=sqrt{frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}geqsqrt{frac{(a^2+b^2+c^2)^3}{3}}=81.$$






                                        share|cite|improve this answer














                                        For $a=b=c=3$ we obtain a value $81$.



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



                                        Indeed, we need to prove that $$a^3+b^3+c^3geq81,$$ which is true because
                                        $$a^3+b^3+c^3-81=sum_{cyc}(a^3-27)=$$
                                        $$=sum_{cyc}left(a^3-27-frac{9}{2}(a^2-9)right)=frac{1}{2}sum_{cyc}(a-3)^2(2a+3)geq0.$$
                                        Also, Holder helps:
                                        $$a^3+b^3+c^3=sqrt{frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}geqsqrt{frac{(a^2+b^2+c^2)^3}{3}}=81.$$







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Nov 16 at 15:21

























                                        answered Nov 16 at 15:14









                                        Michael Rozenberg

                                        94.3k1588183




                                        94.3k1588183












                                        • From which source can I understand holder's inequality with elementary illustrations and exercises. You seem to be good at this.
                                          – Ramanujam Ganit Prashikshan Ke
                                          Nov 16 at 16:55










                                        • @Ramanujam Ganit Prashikshan Ke You can read solutions with tag holder-inequality in this forum. I think it's the best illustration. See also here: math.stackexchange.com/tags/holder-inequality/info
                                          – Michael Rozenberg
                                          Nov 16 at 17:29




















                                        • From which source can I understand holder's inequality with elementary illustrations and exercises. You seem to be good at this.
                                          – Ramanujam Ganit Prashikshan Ke
                                          Nov 16 at 16:55










                                        • @Ramanujam Ganit Prashikshan Ke You can read solutions with tag holder-inequality in this forum. I think it's the best illustration. See also here: math.stackexchange.com/tags/holder-inequality/info
                                          – Michael Rozenberg
                                          Nov 16 at 17:29


















                                        From which source can I understand holder's inequality with elementary illustrations and exercises. You seem to be good at this.
                                        – Ramanujam Ganit Prashikshan Ke
                                        Nov 16 at 16:55




                                        From which source can I understand holder's inequality with elementary illustrations and exercises. You seem to be good at this.
                                        – Ramanujam Ganit Prashikshan Ke
                                        Nov 16 at 16:55












                                        @Ramanujam Ganit Prashikshan Ke You can read solutions with tag holder-inequality in this forum. I think it's the best illustration. See also here: math.stackexchange.com/tags/holder-inequality/info
                                        – Michael Rozenberg
                                        Nov 16 at 17:29






                                        @Ramanujam Ganit Prashikshan Ke You can read solutions with tag holder-inequality in this forum. I think it's the best illustration. See also here: math.stackexchange.com/tags/holder-inequality/info
                                        – Michael Rozenberg
                                        Nov 16 at 17:29




















                                         

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