Prove that $a^2+b^2+c^2geqslantfrac{1}{3}$ given that $agt0, bgt0, cgt0$ and $a+b+c=1$, using existing AM GM...












1












$begingroup$


Using the AM and GM inequality, given that
$agt0, bgt0, cgt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2geqslantfrac{1}{3}$$






algebra-precalculus inequality a.m.-g.m.-inequality







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closed as off-topic by Eevee Trainer, Théophile, Martin R, José Carlos Santos, Kemono Chen Feb 17 at 10:00


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    $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    Feb 17 at 1:42










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    Feb 17 at 1:45






  • 1




    $begingroup$
    Where exactly did you get stuck with that attempt? What stopped you from progressing? And also, please edit your question post with this information as that makes it easier for new readers to catch up (they won't have to sift through comments).
    $endgroup$
    – Arthur
    Feb 17 at 1:50








  • 1




    $begingroup$
    Possible duplicate of Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 ge frac{1}{3}$ holds.
    $endgroup$
    – Martin R
    Feb 17 at 8:31






  • 1




    $begingroup$
    @Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange.
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    – Martin R
    Feb 17 at 17:43
















1












$begingroup$


Using the AM and GM inequality, given that
$agt0, bgt0, cgt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2geqslantfrac{1}{3}$$






algebra-precalculus inequality a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$



closed as off-topic by Eevee Trainer, Théophile, Martin R, José Carlos Santos, Kemono Chen Feb 17 at 10:00


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Théophile, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    Feb 17 at 1:42










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    Feb 17 at 1:45






  • 1




    $begingroup$
    Where exactly did you get stuck with that attempt? What stopped you from progressing? And also, please edit your question post with this information as that makes it easier for new readers to catch up (they won't have to sift through comments).
    $endgroup$
    – Arthur
    Feb 17 at 1:50








  • 1




    $begingroup$
    Possible duplicate of Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 ge frac{1}{3}$ holds.
    $endgroup$
    – Martin R
    Feb 17 at 8:31






  • 1




    $begingroup$
    @Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange.
    $endgroup$
    – Martin R
    Feb 17 at 17:43














1












1








1





$begingroup$


Using the AM and GM inequality, given that
$agt0, bgt0, cgt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2geqslantfrac{1}{3}$$






algebra-precalculus inequality a.m.-g.m.-inequality







share|cite|improve this question











$endgroup$




Using the AM and GM inequality, given that
$agt0, bgt0, cgt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2geqslantfrac{1}{3}$$






algebra-precalculus inequality a.m.-g.m.-inequality




algebra-precalculus inequality a.m.-g.m.-inequality






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share|cite|improve this question













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edited Feb 17 at 8:07









Asaf Karagila♦

306k33438769




306k33438769










asked Feb 17 at 1:38









T. JoelT. Joel

295




295




closed as off-topic by Eevee Trainer, Théophile, Martin R, José Carlos Santos, Kemono Chen Feb 17 at 10:00


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Théophile, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eevee Trainer, Théophile, Martin R, José Carlos Santos, Kemono Chen Feb 17 at 10:00


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Théophile, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    Feb 17 at 1:42










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    Feb 17 at 1:45






  • 1




    $begingroup$
    Where exactly did you get stuck with that attempt? What stopped you from progressing? And also, please edit your question post with this information as that makes it easier for new readers to catch up (they won't have to sift through comments).
    $endgroup$
    – Arthur
    Feb 17 at 1:50








  • 1




    $begingroup$
    Possible duplicate of Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 ge frac{1}{3}$ holds.
    $endgroup$
    – Martin R
    Feb 17 at 8:31






  • 1




    $begingroup$
    @Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange.
    $endgroup$
    – Martin R
    Feb 17 at 17:43














  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    Feb 17 at 1:42










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    Feb 17 at 1:45






  • 1




    $begingroup$
    Where exactly did you get stuck with that attempt? What stopped you from progressing? And also, please edit your question post with this information as that makes it easier for new readers to catch up (they won't have to sift through comments).
    $endgroup$
    – Arthur
    Feb 17 at 1:50








  • 1




    $begingroup$
    Possible duplicate of Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 ge frac{1}{3}$ holds.
    $endgroup$
    – Martin R
    Feb 17 at 8:31






  • 1




    $begingroup$
    @Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange.
    $endgroup$
    – Martin R
    Feb 17 at 17:43








1




1




$begingroup$
What have you tried?
$endgroup$
– Thomas Shelby
Feb 17 at 1:42




$begingroup$
What have you tried?
$endgroup$
– Thomas Shelby
Feb 17 at 1:42












$begingroup$
Using (a+b+c)^2 = 1 but I got stuck
$endgroup$
– T. Joel
Feb 17 at 1:45




$begingroup$
Using (a+b+c)^2 = 1 but I got stuck
$endgroup$
– T. Joel
Feb 17 at 1:45




1




1




$begingroup$
Where exactly did you get stuck with that attempt? What stopped you from progressing? And also, please edit your question post with this information as that makes it easier for new readers to catch up (they won't have to sift through comments).
$endgroup$
– Arthur
Feb 17 at 1:50






$begingroup$
Where exactly did you get stuck with that attempt? What stopped you from progressing? And also, please edit your question post with this information as that makes it easier for new readers to catch up (they won't have to sift through comments).
$endgroup$
– Arthur
Feb 17 at 1:50






1




1




$begingroup$
Possible duplicate of Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 ge frac{1}{3}$ holds.
$endgroup$
– Martin R
Feb 17 at 8:31




$begingroup$
Possible duplicate of Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 ge frac{1}{3}$ holds.
$endgroup$
– Martin R
Feb 17 at 8:31




1




1




$begingroup$
@Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange.
$endgroup$
– Martin R
Feb 17 at 17:43




$begingroup$
@Théophile: More Information here: How to search on this site? and here: Announcing a third-party search engine for Math StackExchange.
$endgroup$
– Martin R
Feb 17 at 17:43










3 Answers
3






active

oldest

votes


















5












$begingroup$

$$a^2+{1over 9} + b^2+{1over 9} + c^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    HINT: You can use your idea of squaring $a+b+c$, but also note that $color{blue}{ab+bc+ca le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)



    One more hint (based on a suggestion from user qsmy): let $x = a^2+b^2+c^2$ and $y = ab+bc+ca$. Squaring both sides of $a+b+c=1$ gives $x+2y=1$, and the blue inequality is $xgeq y$. Can you see it now?






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks!
      $endgroup$
      – T. Joel
      Feb 17 at 2:16










    • $begingroup$
      If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term.
      $endgroup$
      – Minus One-Twelfth
      Feb 17 at 2:24



















    3












    $begingroup$

    In the worst case possible you'd get $$a = b = c = frac{1}{3} Longrightarrow a^2 + b^2 + c^2 = frac{1}{9} + frac{1}{9} + frac{1}{9} = frac{3}{9} geq frac{1}{3} $$



    In the best case possible you'd get $$a = 1, b = c = 0 Longrightarrow 1^2 + 0^2 + 0^2 = 1 geq 1/3 $$



    Therefore the inequality holds. Didn't use the AM-GM inequality, though.






    share|cite|improve this answer











    $endgroup$




















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      $$a^2+{1over 9} + b^2+{1over 9} + c^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.






      share|cite|improve this answer











      $endgroup$


















        5












        $begingroup$

        $$a^2+{1over 9} + b^2+{1over 9} + c^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.






        share|cite|improve this answer











        $endgroup$
















          5












          5








          5





          $begingroup$

          $$a^2+{1over 9} + b^2+{1over 9} + c^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.






          share|cite|improve this answer











          $endgroup$



          $$a^2+{1over 9} + b^2+{1over 9} + c^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 9 at 14:36









          Dr. Mathva

          2,662526




          2,662526










          answered Feb 17 at 2:30









          cr001cr001

          7,854517




          7,854517























              6












              $begingroup$

              HINT: You can use your idea of squaring $a+b+c$, but also note that $color{blue}{ab+bc+ca le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)



              One more hint (based on a suggestion from user qsmy): let $x = a^2+b^2+c^2$ and $y = ab+bc+ca$. Squaring both sides of $a+b+c=1$ gives $x+2y=1$, and the blue inequality is $xgeq y$. Can you see it now?






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks!
                $endgroup$
                – T. Joel
                Feb 17 at 2:16










              • $begingroup$
                If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term.
                $endgroup$
                – Minus One-Twelfth
                Feb 17 at 2:24
















              6












              $begingroup$

              HINT: You can use your idea of squaring $a+b+c$, but also note that $color{blue}{ab+bc+ca le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)



              One more hint (based on a suggestion from user qsmy): let $x = a^2+b^2+c^2$ and $y = ab+bc+ca$. Squaring both sides of $a+b+c=1$ gives $x+2y=1$, and the blue inequality is $xgeq y$. Can you see it now?






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks!
                $endgroup$
                – T. Joel
                Feb 17 at 2:16










              • $begingroup$
                If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term.
                $endgroup$
                – Minus One-Twelfth
                Feb 17 at 2:24














              6












              6








              6





              $begingroup$

              HINT: You can use your idea of squaring $a+b+c$, but also note that $color{blue}{ab+bc+ca le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)



              One more hint (based on a suggestion from user qsmy): let $x = a^2+b^2+c^2$ and $y = ab+bc+ca$. Squaring both sides of $a+b+c=1$ gives $x+2y=1$, and the blue inequality is $xgeq y$. Can you see it now?






              share|cite|improve this answer











              $endgroup$



              HINT: You can use your idea of squaring $a+b+c$, but also note that $color{blue}{ab+bc+ca le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)



              One more hint (based on a suggestion from user qsmy): let $x = a^2+b^2+c^2$ and $y = ab+bc+ca$. Squaring both sides of $a+b+c=1$ gives $x+2y=1$, and the blue inequality is $xgeq y$. Can you see it now?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Feb 17 at 2:31

























              answered Feb 17 at 2:03









              Minus One-TwelfthMinus One-Twelfth

              2,41339




              2,41339








              • 1




                $begingroup$
                I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks!
                $endgroup$
                – T. Joel
                Feb 17 at 2:16










              • $begingroup$
                If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term.
                $endgroup$
                – Minus One-Twelfth
                Feb 17 at 2:24














              • 1




                $begingroup$
                I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks!
                $endgroup$
                – T. Joel
                Feb 17 at 2:16










              • $begingroup$
                If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term.
                $endgroup$
                – Minus One-Twelfth
                Feb 17 at 2:24








              1




              1




              $begingroup$
              I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks!
              $endgroup$
              – T. Joel
              Feb 17 at 2:16




              $begingroup$
              I know the inequality that you stated, but I just can't seem to connect it with my question, please help. Thanks!
              $endgroup$
              – T. Joel
              Feb 17 at 2:16












              $begingroup$
              If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term.
              $endgroup$
              – Minus One-Twelfth
              Feb 17 at 2:24




              $begingroup$
              If you expand $1=(a+b+c)^2$, you should find $ab+bc+ca$ pop up. Apply the blue inequality above to this term.
              $endgroup$
              – Minus One-Twelfth
              Feb 17 at 2:24











              3












              $begingroup$

              In the worst case possible you'd get $$a = b = c = frac{1}{3} Longrightarrow a^2 + b^2 + c^2 = frac{1}{9} + frac{1}{9} + frac{1}{9} = frac{3}{9} geq frac{1}{3} $$



              In the best case possible you'd get $$a = 1, b = c = 0 Longrightarrow 1^2 + 0^2 + 0^2 = 1 geq 1/3 $$



              Therefore the inequality holds. Didn't use the AM-GM inequality, though.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                In the worst case possible you'd get $$a = b = c = frac{1}{3} Longrightarrow a^2 + b^2 + c^2 = frac{1}{9} + frac{1}{9} + frac{1}{9} = frac{3}{9} geq frac{1}{3} $$



                In the best case possible you'd get $$a = 1, b = c = 0 Longrightarrow 1^2 + 0^2 + 0^2 = 1 geq 1/3 $$



                Therefore the inequality holds. Didn't use the AM-GM inequality, though.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  In the worst case possible you'd get $$a = b = c = frac{1}{3} Longrightarrow a^2 + b^2 + c^2 = frac{1}{9} + frac{1}{9} + frac{1}{9} = frac{3}{9} geq frac{1}{3} $$



                  In the best case possible you'd get $$a = 1, b = c = 0 Longrightarrow 1^2 + 0^2 + 0^2 = 1 geq 1/3 $$



                  Therefore the inequality holds. Didn't use the AM-GM inequality, though.






                  share|cite|improve this answer











                  $endgroup$



                  In the worst case possible you'd get $$a = b = c = frac{1}{3} Longrightarrow a^2 + b^2 + c^2 = frac{1}{9} + frac{1}{9} + frac{1}{9} = frac{3}{9} geq frac{1}{3} $$



                  In the best case possible you'd get $$a = 1, b = c = 0 Longrightarrow 1^2 + 0^2 + 0^2 = 1 geq 1/3 $$



                  Therefore the inequality holds. Didn't use the AM-GM inequality, though.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 17 at 2:32

























                  answered Feb 17 at 1:52









                  Victor S.Victor S.

                  31919




                  31919















                      -

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