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
New contributor
$endgroup$
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
New contributor
$endgroup$
$begingroup$
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
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
New contributor
$endgroup$
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
New contributor
New contributor
New contributor
asked 1 hour ago
T. JoelT. Joel
215
215
New contributor
New contributor
$begingroup$
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 |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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?
New contributor
$endgroup$
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
46 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
38 mins ago
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.
$endgroup$
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.
$endgroup$
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
$begingroup$
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?
New contributor
$endgroup$
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
46 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
38 mins 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?
New contributor
$endgroup$
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
46 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
38 mins 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?
New contributor
$endgroup$
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?
New contributor
edited 31 mins ago
New contributor
answered 59 mins ago
Minus One-TwelfthMinus One-Twelfth
6847
6847
New contributor
New contributor
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
46 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
38 mins 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
46 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
38 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
46 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
46 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
38 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
38 mins ago
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
edited 30 mins ago
answered 1 hour ago
Victor S.Victor S.
1348
1348
add a comment |
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
$$a^2+{1over 9} + b^2+{1over 9} + b^2+{1over 9}geq {2over 3}(a+b+c)$$ by AM-GM.
answered 32 mins 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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$begingroup$
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