Prove an inequality, using existing AM GM inequality
$begingroup$
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
asked 1 hour ago
T. JoelT. Joel
265
New contributor
T. Joel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
T. JoelT. Joel
265
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.
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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
asked 1 hour ago
T. JoelT. Joel
265
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
asked 1 hour ago
T. JoelT. Joel
265
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.
asked 1 hour ago
T. JoelT. Joel
265
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.
asked 1 hour ago
T. JoelT. Joel
265
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.
asked 1 hour ago
T. JoelT. Joel
265
asked 1 hour ago
T. JoelT. Joel
265
265
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.
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.
T. Joel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
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
add a comment |
1
$begingroup$
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
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered 1 hour ago
Victor S.Victor S.
1448
$endgroup$
add a comment |
$begingroup$
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?
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
6847
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
add a comment |
$begingroup$
$$a^2+{1over 9} + b^2+{1over 9} + b^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.
answered 1 hour ago
cr001cr001
7,754517
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 1 hour ago
Victor S.Victor S.
1448
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
Victor S.Victor S.
1448
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
Victor S.Victor S.
1448
$endgroup$
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.
answered 1 hour ago
Victor S.Victor S.
1448
edited 1 hour ago
answered 1 hour ago
Victor S.Victor S.
1448
answered 1 hour ago
Victor S.Victor S.
1448
answered 1 hour ago
Victor S.Victor S.
1448
1448
add a comment |
add a comment |
$begingroup$
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?
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
6847
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
add a comment |
$begingroup$
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?
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
6847
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
add a comment |
$begingroup$
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?
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
6847
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?
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
6847
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.
edited 1 hour ago
answered 1 hour ago
Minus One-TwelfthMinus One-Twelfth
6847
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
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.
Check out our Code of Conduct.
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
add a comment |
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
add a comment |
$begingroup$
$$a^2+{1over 9} + b^2+{1over 9} + b^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.
answered 1 hour ago
cr001cr001
7,754517
$endgroup$
add a comment |
$begingroup$
$$a^2+{1over 9} + b^2+{1over 9} + b^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.
answered 1 hour ago
cr001cr001
7,754517
$endgroup$
add a comment |
$begingroup$
$$a^2+{1over 9} + b^2+{1over 9} + b^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.
answered 1 hour ago
cr001cr001
7,754517
$endgroup$
$$a^2+{1over 9} + b^2+{1over 9} + b^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.
answered 1 hour ago
cr001cr001
7,754517
answered 1 hour ago
cr001cr001
7,754517
answered 1 hour ago
cr001cr001
7,754517
answered 1 hour ago
cr001cr001
7,754517
7,754517
add a comment |
add a comment |
T. Joel is a new contributor. Be nice, and check out our Code of Conduct.
T. Joel is a new contributor. Be nice, and check out our Code of Conduct.
T. Joel is a new contributor. Be nice, and check out our Code of Conduct.
T. Joel is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
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