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
    1 hour ago










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










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









T. JoelT. Joel

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  • $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    1 hour ago










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




















  • $begingroup$
    What have you tried?
    $endgroup$
    – Thomas Shelby
    1 hour ago










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


















$begingroup$
What have you tried?
$endgroup$
– Thomas Shelby
1 hour ago




$begingroup$
What have you tried?
$endgroup$
– Thomas Shelby
1 hour ago












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




$begingroup$
Using (a+b+c)^2 = 1 but I got stuck
$endgroup$
– T. Joel
1 hour 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
1 hour 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
1 hour 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
    1 hour 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
    1 hour 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|improve this answer









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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
        1 hour 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
        1 hour 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
        1 hour 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
        1 hour 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 55 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 1 hour ago









      Minus One-TwelfthMinus One-Twelfth

      6847




      6847




      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
        1 hour 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
        1 hour 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
        1 hour 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
        1 hour 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
      1 hour 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
      1 hour 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
      1 hour 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
      1 hour 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 54 mins ago

























          answered 1 hour 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|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|improve this answer









                $endgroup$
















                  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|improve this answer









                  $endgroup$



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







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 55 mins ago









                  cr001cr001

                  7,754517




                  7,754517






















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