Prove an inequality, using existing AM GM inequality












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 proof-verification a.m.-g.m.-inequality







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  • $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    56 mins ago










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    53 mins ago










  • $begingroup$
    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
    48 mins ago


















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 proof-verification a.m.-g.m.-inequality







share|cite|improve this question







New contributor




T. Joel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    56 mins ago










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    53 mins ago










  • $begingroup$
    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
    48 mins ago
















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 proof-verification a.m.-g.m.-inequality







share|cite|improve this question







New contributor




T. Joel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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 proof-verification a.m.-g.m.-inequality




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






share|cite|improve this question







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T. Joel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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T. Joel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 59 mins ago









T. JoelT. Joel

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  • $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    56 mins ago










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    53 mins ago










  • $begingroup$
    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
    48 mins ago




















  • $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    56 mins ago










  • $begingroup$
    Using (a+b+c)^2 = 1 but I got stuck
    $endgroup$
    – T. Joel
    53 mins ago










  • $begingroup$
    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
    48 mins ago


















$begingroup$
What have you tried?
$endgroup$
– Thomas Shelby
56 mins ago




$begingroup$
What have you tried?
$endgroup$
– Thomas Shelby
56 mins ago












$begingroup$
Using (a+b+c)^2 = 1 but I got stuck
$endgroup$
– T. Joel
53 mins ago




$begingroup$
Using (a+b+c)^2 = 1 but I got stuck
$endgroup$
– T. Joel
53 mins ago












$begingroup$
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
48 mins ago






$begingroup$
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
48 mins ago












3 Answers
3






active

oldest

votes


















2












$begingroup$

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










New contributor




Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$









  • 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
    21 mins ago










  • $begingroup$
    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
    14 mins ago



















2












$begingroup$

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











$endgroup$





















    0












    $begingroup$

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






    share|cite









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      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










      New contributor




      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$









      • 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
        21 mins ago










      • $begingroup$
        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
        14 mins ago
















      2












      $begingroup$

      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










      New contributor




      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$









      • 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
        21 mins ago










      • $begingroup$
        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
        14 mins ago














      2












      2








      2





      $begingroup$

      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










      New contributor




      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$



      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










      New contributor




      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this answer



      share|cite|improve this answer








      edited 7 mins ago





















      New contributor




      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      answered 35 mins ago









      Minus One-TwelfthMinus One-Twelfth

      6497




      6497




      New contributor




      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Minus One-Twelfth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      • 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
        21 mins ago










      • $begingroup$
        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
        14 mins ago














      • 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
        21 mins ago










      • $begingroup$
        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
        14 mins ago








      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
      21 mins ago




      $begingroup$
      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
      21 mins ago












      $begingroup$
      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
      14 mins ago




      $begingroup$
      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
      14 mins ago











      2












      $begingroup$

      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











      $endgroup$


















        2












        $begingroup$

        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











        $endgroup$
















          2












          2








          2





          $begingroup$

          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 6 mins ago

























          answered 45 mins ago









          Victor S.Victor S.

          1348




          1348























              0












              $begingroup$

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






              share|cite









              $endgroup$


















                0












                $begingroup$

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






                share|cite









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

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






                  share|cite









                  $endgroup$



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







                  share|cite












                  share|cite



                  share|cite










                  answered 8 mins ago









                  cr001cr001

                  7,754517




                  7,754517






















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