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Application of Chebyschev inequality
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I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the term to its right to get
$$fraca+b+c3cdot fraca+b+c3geq fracca+b^2+ca3$$
My problem
I would've wanted to get $dfracab+bc+ca3$ on the right instead. Did I apply the inequality wrong, or does it follow that $dfracab+bc+ca3$ is less than or equal to $dfraca+b+c3cdot dfraca+b+c3$?
inequality proof-writing rearrangement-inequality
edited Jul 31 at 18:57
Michael Rozenberg
87.5k1577179
asked Jul 31 at 14:30
john fowles
1,088817
add a comment |Â
up vote
4
down vote
favorite
I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the term to its right to get
$$fraca+b+c3cdot fraca+b+c3geq fracca+b^2+ca3$$
My problem
I would've wanted to get $dfracab+bc+ca3$ on the right instead. Did I apply the inequality wrong, or does it follow that $dfracab+bc+ca3$ is less than or equal to $dfraca+b+c3cdot dfraca+b+c3$?
inequality proof-writing rearrangement-inequality
edited Jul 31 at 18:57
Michael Rozenberg
87.5k1577179
asked Jul 31 at 14:30
john fowles
1,088817
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the term to its right to get
$$fraca+b+c3cdot fraca+b+c3geq fracca+b^2+ca3$$
My problem
I would've wanted to get $dfracab+bc+ca3$ on the right instead. Did I apply the inequality wrong, or does it follow that $dfracab+bc+ca3$ is less than or equal to $dfraca+b+c3cdot dfraca+b+c3$?
inequality proof-writing rearrangement-inequality
edited Jul 31 at 18:57
Michael Rozenberg
87.5k1577179
asked Jul 31 at 14:30
john fowles
1,088817
I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the term to its right to get
$$fraca+b+c3cdot fraca+b+c3geq fracca+b^2+ca3$$
My problem
I would've wanted to get $dfracab+bc+ca3$ on the right instead. Did I apply the inequality wrong, or does it follow that $dfracab+bc+ca3$ is less than or equal to $dfraca+b+c3cdot dfraca+b+c3$?
inequality proof-writing rearrangement-inequality
edited Jul 31 at 18:57
Michael Rozenberg
87.5k1577179
asked Jul 31 at 14:30
john fowles
1,088817
edited Jul 31 at 18:57
Michael Rozenberg
87.5k1577179
edited Jul 31 at 18:57
Michael Rozenberg
87.5k1577179
edited Jul 31 at 18:57
Michael Rozenberg
87.5k1577179
87.5k1577179
asked Jul 31 at 14:30
john fowles
1,088817
asked Jul 31 at 14:30
john fowles
1,088817
asked Jul 31 at 14:30
john fowles
1,088817
1,088817
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain:
$$3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)$$or
$$3(a^2+b^2+c^2)geq(a+b+c)^2$$ or
$$3(a^2+b^2+c^2+2(ab+ac+bc))geq(a+b+c)^2+2(ab+ac+bc)$$ or
$$(a+b+c)^2geq3(ab+ac+bc)$$ or
$$fraca+b+c3geqsqrtfracab+ac+bc3.$$
By the same way we obtain:
$$3(abcdot ab+accdot ac+bccdot bc)geq (ab+ac+bc)(ab+ac+bc)$$ or
$$3(a^2b^2+a^2c^2+b^2c^2)geq(ab+ac+bc)^2$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c).$$
Now, $(ab,ac,bc)$ and $(c,b,a)$ they are opposite ordered.
Thus, by Chebyshov again we obtain:
$$(ab+ac+bc)^3geq3abc(a+b+c)(ab+ac+bc)=3abc(c+b+a)(ab+ac+bc)geq$$
$$geq3abccdot3(ccdot ab+bcdot ac+acdot bc)=27a^2b^2c^2,$$
which gives
$$sqrtfracab+ac+bc3geqsqrt[3]abc.$$
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
add a comment |Â
up vote
1
down vote
As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 ge ab + bc + ca$.
The last equality is true, as we have that $a^2 + b^2 ge 2ab$, $b^2 + c^2 ge 2bc$ and $c^2 + a^2 ge 2ac$. Add the inequalities to get the final answer.
answered Jul 31 at 14:47
Stefan4024
27.8k52974
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain:
$$3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)$$or
$$3(a^2+b^2+c^2)geq(a+b+c)^2$$ or
$$3(a^2+b^2+c^2+2(ab+ac+bc))geq(a+b+c)^2+2(ab+ac+bc)$$ or
$$(a+b+c)^2geq3(ab+ac+bc)$$ or
$$fraca+b+c3geqsqrtfracab+ac+bc3.$$
By the same way we obtain:
$$3(abcdot ab+accdot ac+bccdot bc)geq (ab+ac+bc)(ab+ac+bc)$$ or
$$3(a^2b^2+a^2c^2+b^2c^2)geq(ab+ac+bc)^2$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c).$$
Now, $(ab,ac,bc)$ and $(c,b,a)$ they are opposite ordered.
Thus, by Chebyshov again we obtain:
$$(ab+ac+bc)^3geq3abc(a+b+c)(ab+ac+bc)=3abc(c+b+a)(ab+ac+bc)geq$$
$$geq3abccdot3(ccdot ab+bcdot ac+acdot bc)=27a^2b^2c^2,$$
which gives
$$sqrtfracab+ac+bc3geqsqrt[3]abc.$$
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
add a comment |Â
up vote
2
down vote
accepted
Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain:
$$3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)$$or
$$3(a^2+b^2+c^2)geq(a+b+c)^2$$ or
$$3(a^2+b^2+c^2+2(ab+ac+bc))geq(a+b+c)^2+2(ab+ac+bc)$$ or
$$(a+b+c)^2geq3(ab+ac+bc)$$ or
$$fraca+b+c3geqsqrtfracab+ac+bc3.$$
By the same way we obtain:
$$3(abcdot ab+accdot ac+bccdot bc)geq (ab+ac+bc)(ab+ac+bc)$$ or
$$3(a^2b^2+a^2c^2+b^2c^2)geq(ab+ac+bc)^2$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c).$$
Now, $(ab,ac,bc)$ and $(c,b,a)$ they are opposite ordered.
Thus, by Chebyshov again we obtain:
$$(ab+ac+bc)^3geq3abc(a+b+c)(ab+ac+bc)=3abc(c+b+a)(ab+ac+bc)geq$$
$$geq3abccdot3(ccdot ab+bcdot ac+acdot bc)=27a^2b^2c^2,$$
which gives
$$sqrtfracab+ac+bc3geqsqrt[3]abc.$$
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain:
$$3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)$$or
$$3(a^2+b^2+c^2)geq(a+b+c)^2$$ or
$$3(a^2+b^2+c^2+2(ab+ac+bc))geq(a+b+c)^2+2(ab+ac+bc)$$ or
$$(a+b+c)^2geq3(ab+ac+bc)$$ or
$$fraca+b+c3geqsqrtfracab+ac+bc3.$$
By the same way we obtain:
$$3(abcdot ab+accdot ac+bccdot bc)geq (ab+ac+bc)(ab+ac+bc)$$ or
$$3(a^2b^2+a^2c^2+b^2c^2)geq(ab+ac+bc)^2$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c).$$
Now, $(ab,ac,bc)$ and $(c,b,a)$ they are opposite ordered.
Thus, by Chebyshov again we obtain:
$$(ab+ac+bc)^3geq3abc(a+b+c)(ab+ac+bc)=3abc(c+b+a)(ab+ac+bc)geq$$
$$geq3abccdot3(ccdot ab+bcdot ac+acdot bc)=27a^2b^2c^2,$$
which gives
$$sqrtfracab+ac+bc3geqsqrt[3]abc.$$
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
Firstly, your inequality is wrong.
Try $a=1$, $b=-1$ and $c=0$.
For non-negatives $a$, $b$ and $c$ we see that $(a,b,c)$ and $(a,b,c)$ they are the same ordered.
This, by Chebyshov we obtain:
$$3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)$$or
$$3(a^2+b^2+c^2)geq(a+b+c)^2$$ or
$$3(a^2+b^2+c^2+2(ab+ac+bc))geq(a+b+c)^2+2(ab+ac+bc)$$ or
$$(a+b+c)^2geq3(ab+ac+bc)$$ or
$$fraca+b+c3geqsqrtfracab+ac+bc3.$$
By the same way we obtain:
$$3(abcdot ab+accdot ac+bccdot bc)geq (ab+ac+bc)(ab+ac+bc)$$ or
$$3(a^2b^2+a^2c^2+b^2c^2)geq(ab+ac+bc)^2$$ or
$$(ab+ac+bc)^2geq3abc(a+b+c).$$
Now, $(ab,ac,bc)$ and $(c,b,a)$ they are opposite ordered.
Thus, by Chebyshov again we obtain:
$$(ab+ac+bc)^3geq3abc(a+b+c)(ab+ac+bc)=3abc(c+b+a)(ab+ac+bc)geq$$
$$geq3abccdot3(ccdot ab+bcdot ac+acdot bc)=27a^2b^2c^2,$$
which gives
$$sqrtfracab+ac+bc3geqsqrt[3]abc.$$
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
answered Jul 31 at 18:55
Michael Rozenberg
87.5k1577179
87.5k1577179
add a comment |Â
add a comment |Â
up vote
1
down vote
As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 ge ab + bc + ca$.
The last equality is true, as we have that $a^2 + b^2 ge 2ab$, $b^2 + c^2 ge 2bc$ and $c^2 + a^2 ge 2ac$. Add the inequalities to get the final answer.
answered Jul 31 at 14:47
Stefan4024
27.8k52974
add a comment |Â
up vote
1
down vote
As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 ge ab + bc + ca$.
The last equality is true, as we have that $a^2 + b^2 ge 2ab$, $b^2 + c^2 ge 2bc$ and $c^2 + a^2 ge 2ac$. Add the inequalities to get the final answer.
answered Jul 31 at 14:47
Stefan4024
27.8k52974
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 ge ab + bc + ca$.
The last equality is true, as we have that $a^2 + b^2 ge 2ab$, $b^2 + c^2 ge 2bc$ and $c^2 + a^2 ge 2ac$. Add the inequalities to get the final answer.
answered Jul 31 at 14:47
Stefan4024
27.8k52974
As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 ge ab + bc + ca$.
The last equality is true, as we have that $a^2 + b^2 ge 2ab$, $b^2 + c^2 ge 2bc$ and $c^2 + a^2 ge 2ac$. Add the inequalities to get the final answer.
answered Jul 31 at 14:47
Stefan4024
27.8k52974
answered Jul 31 at 14:47
Stefan4024
27.8k52974
answered Jul 31 at 14:47
Stefan4024
27.8k52974
answered Jul 31 at 14:47
Stefan4024
27.8k52974
27.8k52974
add a comment |Â
add a comment |Â
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