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Prove the inequality using Chebyshev's Inequality
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If $a,b,c in(0,infty)$, then prove that: $$9(a^3+b^3+c^3)ge(a+b+c)^3$$
I was trying to prove this inequality using Chebyshev's Inequality and assuming $age b ge c$ but to no avail.
Can please someone help me in proving it using Chebyshev's Inequality?
algebra-precalculus inequality symmetric-polynomials holder-inequality rearrangement-inequality
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
asked Sep 9 at 15:38
Vivek
1208
add a comment |Â
up vote
3
down vote
favorite
If $a,b,c in(0,infty)$, then prove that: $$9(a^3+b^3+c^3)ge(a+b+c)^3$$
I was trying to prove this inequality using Chebyshev's Inequality and assuming $age b ge c$ but to no avail.
Can please someone help me in proving it using Chebyshev's Inequality?
algebra-precalculus inequality symmetric-polynomials holder-inequality rearrangement-inequality
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
asked Sep 9 at 15:38
Vivek
1208
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $a,b,c in(0,infty)$, then prove that: $$9(a^3+b^3+c^3)ge(a+b+c)^3$$
I was trying to prove this inequality using Chebyshev's Inequality and assuming $age b ge c$ but to no avail.
Can please someone help me in proving it using Chebyshev's Inequality?
algebra-precalculus inequality symmetric-polynomials holder-inequality rearrangement-inequality
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
asked Sep 9 at 15:38
Vivek
1208
If $a,b,c in(0,infty)$, then prove that: $$9(a^3+b^3+c^3)ge(a+b+c)^3$$
I was trying to prove this inequality using Chebyshev's Inequality and assuming $age b ge c$ but to no avail.
Can please someone help me in proving it using Chebyshev's Inequality?
algebra-precalculus inequality symmetric-polynomials holder-inequality rearrangement-inequality
algebra-precalculus inequality symmetric-polynomials holder-inequality rearrangement-inequality
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
asked Sep 9 at 15:38
Vivek
1208
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
asked Sep 9 at 15:38
Vivek
1208
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
edited Sep 9 at 15:54
Michael Rozenberg
89.2k1582180
89.2k1582180
asked Sep 9 at 15:38
Vivek
1208
asked Sep 9 at 15:38
Vivek
1208
asked Sep 9 at 15:38
Vivek
1208
1208
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Since $(a,b,c)$ and $(a^2,b^2,c^2)$ they are the same ordered, by Chebyshov twice we ontain:
$$9(a^3+b^3+c^3)=3cdot3(a^3+b^3+c^3)geq3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3.$$
Also, by Holder
$$(1+1+1)^2(a^3+b^3+c^3)geqleft(sqrt[3]1^2a^3+sqrt[3]1^2b^3+sqrt[3]1^2c^3right)^3=(a+b+c)^3.$$
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
How $3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3$?
– Vivek
Sep 9 at 15:50
2
Because by Chebyshov $3(a^2+b^2+c^2)=3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)=(a+b+c)^2.$
– Michael Rozenberg
Sep 9 at 15:52
Ohh..Clear.. Thanks!!
– Vivek
Sep 9 at 15:54
You are welcome!
– Michael Rozenberg
Sep 9 at 15:55
And here I was looking for a fancy probability distribution to solve it... ;p
– Quintec
2 days ago
add a comment |Â
up vote
4
down vote
Recall that by the generalized mean inequality
$$sqrt[3]fraca^3+b^3+c^33ge fraca+b+c3$$
and the result directly follows.
answered Sep 9 at 15:42
gimusi
71.4k73786
Can't this be done by using Chebyshev's Inequality?
– Vivek
Sep 9 at 15:46
It doesn’t seem convenient in that case. Are you forced to use that?
– gimusi
Sep 9 at 15:49
Yes..............
– Vivek
Sep 9 at 15:52
I see now that Michael has solved in that way!
– gimusi
Sep 9 at 15:52
1
Yes... By the way, thanks to you too. Your method was great too.
– Vivek
Sep 9 at 15:55
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Since $(a,b,c)$ and $(a^2,b^2,c^2)$ they are the same ordered, by Chebyshov twice we ontain:
$$9(a^3+b^3+c^3)=3cdot3(a^3+b^3+c^3)geq3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3.$$
Also, by Holder
$$(1+1+1)^2(a^3+b^3+c^3)geqleft(sqrt[3]1^2a^3+sqrt[3]1^2b^3+sqrt[3]1^2c^3right)^3=(a+b+c)^3.$$
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
How $3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3$?
– Vivek
Sep 9 at 15:50
2
Because by Chebyshov $3(a^2+b^2+c^2)=3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)=(a+b+c)^2.$
– Michael Rozenberg
Sep 9 at 15:52
Ohh..Clear.. Thanks!!
– Vivek
Sep 9 at 15:54
You are welcome!
– Michael Rozenberg
Sep 9 at 15:55
And here I was looking for a fancy probability distribution to solve it... ;p
– Quintec
2 days ago
add a comment |Â
up vote
6
down vote
accepted
Since $(a,b,c)$ and $(a^2,b^2,c^2)$ they are the same ordered, by Chebyshov twice we ontain:
$$9(a^3+b^3+c^3)=3cdot3(a^3+b^3+c^3)geq3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3.$$
Also, by Holder
$$(1+1+1)^2(a^3+b^3+c^3)geqleft(sqrt[3]1^2a^3+sqrt[3]1^2b^3+sqrt[3]1^2c^3right)^3=(a+b+c)^3.$$
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
How $3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3$?
– Vivek
Sep 9 at 15:50
2
Because by Chebyshov $3(a^2+b^2+c^2)=3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)=(a+b+c)^2.$
– Michael Rozenberg
Sep 9 at 15:52
Ohh..Clear.. Thanks!!
– Vivek
Sep 9 at 15:54
You are welcome!
– Michael Rozenberg
Sep 9 at 15:55
And here I was looking for a fancy probability distribution to solve it... ;p
– Quintec
2 days ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Since $(a,b,c)$ and $(a^2,b^2,c^2)$ they are the same ordered, by Chebyshov twice we ontain:
$$9(a^3+b^3+c^3)=3cdot3(a^3+b^3+c^3)geq3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3.$$
Also, by Holder
$$(1+1+1)^2(a^3+b^3+c^3)geqleft(sqrt[3]1^2a^3+sqrt[3]1^2b^3+sqrt[3]1^2c^3right)^3=(a+b+c)^3.$$
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
Since $(a,b,c)$ and $(a^2,b^2,c^2)$ they are the same ordered, by Chebyshov twice we ontain:
$$9(a^3+b^3+c^3)=3cdot3(a^3+b^3+c^3)geq3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3.$$
Also, by Holder
$$(1+1+1)^2(a^3+b^3+c^3)geqleft(sqrt[3]1^2a^3+sqrt[3]1^2b^3+sqrt[3]1^2c^3right)^3=(a+b+c)^3.$$
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
answered Sep 9 at 15:44
Michael Rozenberg
89.2k1582180
89.2k1582180
How $3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3$?
– Vivek
Sep 9 at 15:50
2
Because by Chebyshov $3(a^2+b^2+c^2)=3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)=(a+b+c)^2.$
– Michael Rozenberg
Sep 9 at 15:52
Ohh..Clear.. Thanks!!
– Vivek
Sep 9 at 15:54
You are welcome!
– Michael Rozenberg
Sep 9 at 15:55
And here I was looking for a fancy probability distribution to solve it... ;p
– Quintec
2 days ago
add a comment |Â
How $3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3$?
– Vivek
Sep 9 at 15:50
2
Because by Chebyshov $3(a^2+b^2+c^2)=3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)=(a+b+c)^2.$
– Michael Rozenberg
Sep 9 at 15:52
Ohh..Clear.. Thanks!!
– Vivek
Sep 9 at 15:54
You are welcome!
– Michael Rozenberg
Sep 9 at 15:55
And here I was looking for a fancy probability distribution to solve it... ;p
– Quintec
2 days ago
How $3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3$?
– Vivek
Sep 9 at 15:50
How $3(a+b+c)(a^2+b^2+c^2)geq(a+b+c)^3$?
– Vivek
Sep 9 at 15:50
2
2
Because by Chebyshov $3(a^2+b^2+c^2)=3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)=(a+b+c)^2.$
– Michael Rozenberg
Sep 9 at 15:52
Because by Chebyshov $3(a^2+b^2+c^2)=3(acdot a+bcdot b+ccdot c)geq(a+b+c)(a+b+c)=(a+b+c)^2.$
– Michael Rozenberg
Sep 9 at 15:52
Ohh..Clear.. Thanks!!
– Vivek
Sep 9 at 15:54
Ohh..Clear.. Thanks!!
– Vivek
Sep 9 at 15:54
You are welcome!
– Michael Rozenberg
Sep 9 at 15:55
You are welcome!
– Michael Rozenberg
Sep 9 at 15:55
And here I was looking for a fancy probability distribution to solve it... ;p
– Quintec
2 days ago
And here I was looking for a fancy probability distribution to solve it... ;p
– Quintec
2 days ago
add a comment |Â
up vote
4
down vote
Recall that by the generalized mean inequality
$$sqrt[3]fraca^3+b^3+c^33ge fraca+b+c3$$
and the result directly follows.
answered Sep 9 at 15:42
gimusi
71.4k73786
Can't this be done by using Chebyshev's Inequality?
– Vivek
Sep 9 at 15:46
It doesn’t seem convenient in that case. Are you forced to use that?
– gimusi
Sep 9 at 15:49
Yes..............
– Vivek
Sep 9 at 15:52
I see now that Michael has solved in that way!
– gimusi
Sep 9 at 15:52
1
Yes... By the way, thanks to you too. Your method was great too.
– Vivek
Sep 9 at 15:55
 |Â
show 1 more comment
up vote
4
down vote
Recall that by the generalized mean inequality
$$sqrt[3]fraca^3+b^3+c^33ge fraca+b+c3$$
and the result directly follows.
answered Sep 9 at 15:42
gimusi
71.4k73786
Can't this be done by using Chebyshev's Inequality?
– Vivek
Sep 9 at 15:46
It doesn’t seem convenient in that case. Are you forced to use that?
– gimusi
Sep 9 at 15:49
Yes..............
– Vivek
Sep 9 at 15:52
I see now that Michael has solved in that way!
– gimusi
Sep 9 at 15:52
1
Yes... By the way, thanks to you too. Your method was great too.
– Vivek
Sep 9 at 15:55
 |Â
show 1 more comment
up vote
4
down vote
up vote
4
down vote
Recall that by the generalized mean inequality
$$sqrt[3]fraca^3+b^3+c^33ge fraca+b+c3$$
and the result directly follows.
answered Sep 9 at 15:42
gimusi
71.4k73786
Recall that by the generalized mean inequality
$$sqrt[3]fraca^3+b^3+c^33ge fraca+b+c3$$
and the result directly follows.
answered Sep 9 at 15:42
gimusi
71.4k73786
answered Sep 9 at 15:42
gimusi
71.4k73786
answered Sep 9 at 15:42
gimusi
71.4k73786
answered Sep 9 at 15:42
gimusi
71.4k73786
71.4k73786
Can't this be done by using Chebyshev's Inequality?
– Vivek
Sep 9 at 15:46
It doesn’t seem convenient in that case. Are you forced to use that?
– gimusi
Sep 9 at 15:49
Yes..............
– Vivek
Sep 9 at 15:52
I see now that Michael has solved in that way!
– gimusi
Sep 9 at 15:52
1
Yes... By the way, thanks to you too. Your method was great too.
– Vivek
Sep 9 at 15:55
 |Â
show 1 more comment
Can't this be done by using Chebyshev's Inequality?
– Vivek
Sep 9 at 15:46
It doesn’t seem convenient in that case. Are you forced to use that?
– gimusi
Sep 9 at 15:49
Yes..............
– Vivek
Sep 9 at 15:52
I see now that Michael has solved in that way!
– gimusi
Sep 9 at 15:52
1
Yes... By the way, thanks to you too. Your method was great too.
– Vivek
Sep 9 at 15:55
Can't this be done by using Chebyshev's Inequality?
– Vivek
Sep 9 at 15:46
Can't this be done by using Chebyshev's Inequality?
– Vivek
Sep 9 at 15:46
It doesn’t seem convenient in that case. Are you forced to use that?
– gimusi
Sep 9 at 15:49
It doesn’t seem convenient in that case. Are you forced to use that?
– gimusi
Sep 9 at 15:49
Yes..............
– Vivek
Sep 9 at 15:52
Yes..............
– Vivek
Sep 9 at 15:52
I see now that Michael has solved in that way!
– gimusi
Sep 9 at 15:52
I see now that Michael has solved in that way!
– gimusi
Sep 9 at 15:52
1
1
Yes... By the way, thanks to you too. Your method was great too.
– Vivek
Sep 9 at 15:55
Yes... By the way, thanks to you too. Your method was great too.
– Vivek
Sep 9 at 15:55
 |Â
show 1 more comment
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