Find least upper bound of ${a^{2018} + b^{2018} + c^{2018} | a + b + c = 1, a, b, c > 0}$
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1
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Find least upper bound of ${a^{2018} + b^{2018} + c^{2018} | a + b + c = 1, a, b, c > 0 }$. I tried power mean inequality, but could only find greatest lower bound.
inequality karamata-inequality
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Nov 17 at 11:46
J. Abraham
486313
add a comment |
up vote
1
down vote
favorite
Find least upper bound of ${a^{2018} + b^{2018} + c^{2018} | a + b + c = 1, a, b, c > 0 }$. I tried power mean inequality, but could only find greatest lower bound.
inequality karamata-inequality
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Nov 17 at 11:46
J. Abraham
486313
The numbers are positive reals.
– J. Abraham
Nov 17 at 11:52
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find least upper bound of ${a^{2018} + b^{2018} + c^{2018} | a + b + c = 1, a, b, c > 0 }$. I tried power mean inequality, but could only find greatest lower bound.
inequality karamata-inequality
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Nov 17 at 11:46
J. Abraham
486313
Find least upper bound of ${a^{2018} + b^{2018} + c^{2018} | a + b + c = 1, a, b, c > 0 }$. I tried power mean inequality, but could only find greatest lower bound.
inequality karamata-inequality
inequality karamata-inequality
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Nov 17 at 11:46
J. Abraham
486313
edited 2 days ago
Michael Rozenberg
94.2k1588183
asked Nov 17 at 11:46
J. Abraham
486313
edited 2 days ago
Michael Rozenberg
94.2k1588183
edited 2 days ago
Michael Rozenberg
94.2k1588183
edited 2 days ago
Michael Rozenberg
94.2k1588183
94.2k1588183
asked Nov 17 at 11:46
J. Abraham
486313
asked Nov 17 at 11:46
J. Abraham
486313
asked Nov 17 at 11:46
J. Abraham
486313
486313
The numbers are positive reals.
– J. Abraham
Nov 17 at 11:52
add a comment |
The numbers are positive reals.
– J. Abraham
Nov 17 at 11:52
The numbers are positive reals.
– J. Abraham
Nov 17 at 11:52
The numbers are positive reals.
– J. Abraham
Nov 17 at 11:52
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Let $f(x)=x^{2018},$ where $xgeq0$ and $ageq bgeq cgeq0.$
Thus, since $(a+b+c,0,0)succ(a,b,c)$ and $f$ is a convex function, by Karamata we obtain:
$$1=(a+b+c)^{2018}+0^{2018}+0^{2018}geq a^{2018}+b^{2018}+c^{2018}.$$
The equality occurs for $a=1$ and $b=c=0$.
In our case our variables are positives, which says that $1$ is a supremum.
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
add a comment |
up vote
5
down vote
Well, $0<a,b,c<1$, so $a^{2018}<a$ etc., therefore $a^{2018}+b^{2018}+c^{2018}<a+b+c=1$. I reckon one can get close to $1$.
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
and a lower bound ?
– Henno Brandsma
Nov 17 at 12:00
Can you prove that it is a least upper bound?
– J. Abraham
Nov 17 at 12:05
@J.Abraham All numbers in the set are $le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g.
– Henno Brandsma
Nov 17 at 12:17
@Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$.
– Sambo
Nov 17 at 12:38
1
@Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$.
– Henno Brandsma
Nov 17 at 12:40
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $f(x)=x^{2018},$ where $xgeq0$ and $ageq bgeq cgeq0.$
Thus, since $(a+b+c,0,0)succ(a,b,c)$ and $f$ is a convex function, by Karamata we obtain:
$$1=(a+b+c)^{2018}+0^{2018}+0^{2018}geq a^{2018}+b^{2018}+c^{2018}.$$
The equality occurs for $a=1$ and $b=c=0$.
In our case our variables are positives, which says that $1$ is a supremum.
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
add a comment |
up vote
2
down vote
accepted
Let $f(x)=x^{2018},$ where $xgeq0$ and $ageq bgeq cgeq0.$
Thus, since $(a+b+c,0,0)succ(a,b,c)$ and $f$ is a convex function, by Karamata we obtain:
$$1=(a+b+c)^{2018}+0^{2018}+0^{2018}geq a^{2018}+b^{2018}+c^{2018}.$$
The equality occurs for $a=1$ and $b=c=0$.
In our case our variables are positives, which says that $1$ is a supremum.
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $f(x)=x^{2018},$ where $xgeq0$ and $ageq bgeq cgeq0.$
Thus, since $(a+b+c,0,0)succ(a,b,c)$ and $f$ is a convex function, by Karamata we obtain:
$$1=(a+b+c)^{2018}+0^{2018}+0^{2018}geq a^{2018}+b^{2018}+c^{2018}.$$
The equality occurs for $a=1$ and $b=c=0$.
In our case our variables are positives, which says that $1$ is a supremum.
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
Let $f(x)=x^{2018},$ where $xgeq0$ and $ageq bgeq cgeq0.$
Thus, since $(a+b+c,0,0)succ(a,b,c)$ and $f$ is a convex function, by Karamata we obtain:
$$1=(a+b+c)^{2018}+0^{2018}+0^{2018}geq a^{2018}+b^{2018}+c^{2018}.$$
The equality occurs for $a=1$ and $b=c=0$.
In our case our variables are positives, which says that $1$ is a supremum.
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
answered Nov 17 at 12:10
Michael Rozenberg
94.2k1588183
94.2k1588183
add a comment |
add a comment |
up vote
5
down vote
Well, $0<a,b,c<1$, so $a^{2018}<a$ etc., therefore $a^{2018}+b^{2018}+c^{2018}<a+b+c=1$. I reckon one can get close to $1$.
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
and a lower bound ?
– Henno Brandsma
Nov 17 at 12:00
Can you prove that it is a least upper bound?
– J. Abraham
Nov 17 at 12:05
@J.Abraham All numbers in the set are $le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g.
– Henno Brandsma
Nov 17 at 12:17
@Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$.
– Sambo
Nov 17 at 12:38
1
@Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$.
– Henno Brandsma
Nov 17 at 12:40
add a comment |
up vote
5
down vote
Well, $0<a,b,c<1$, so $a^{2018}<a$ etc., therefore $a^{2018}+b^{2018}+c^{2018}<a+b+c=1$. I reckon one can get close to $1$.
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
and a lower bound ?
– Henno Brandsma
Nov 17 at 12:00
Can you prove that it is a least upper bound?
– J. Abraham
Nov 17 at 12:05
@J.Abraham All numbers in the set are $le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g.
– Henno Brandsma
Nov 17 at 12:17
@Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$.
– Sambo
Nov 17 at 12:38
1
@Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$.
– Henno Brandsma
Nov 17 at 12:40
add a comment |
up vote
5
down vote
up vote
5
down vote
Well, $0<a,b,c<1$, so $a^{2018}<a$ etc., therefore $a^{2018}+b^{2018}+c^{2018}<a+b+c=1$. I reckon one can get close to $1$.
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
Well, $0<a,b,c<1$, so $a^{2018}<a$ etc., therefore $a^{2018}+b^{2018}+c^{2018}<a+b+c=1$. I reckon one can get close to $1$.
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
answered Nov 17 at 11:54
Lord Shark the Unknown
97.1k958128
97.1k958128
and a lower bound ?
– Henno Brandsma
Nov 17 at 12:00
Can you prove that it is a least upper bound?
– J. Abraham
Nov 17 at 12:05
@J.Abraham All numbers in the set are $le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g.
– Henno Brandsma
Nov 17 at 12:17
@Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$.
– Sambo
Nov 17 at 12:38
1
@Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$.
– Henno Brandsma
Nov 17 at 12:40
add a comment |
and a lower bound ?
– Henno Brandsma
Nov 17 at 12:00
Can you prove that it is a least upper bound?
– J. Abraham
Nov 17 at 12:05
@J.Abraham All numbers in the set are $le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g.
– Henno Brandsma
Nov 17 at 12:17
@Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$.
– Sambo
Nov 17 at 12:38
1
@Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$.
– Henno Brandsma
Nov 17 at 12:40
and a lower bound ?
– Henno Brandsma
Nov 17 at 12:00
and a lower bound ?
– Henno Brandsma
Nov 17 at 12:00
Can you prove that it is a least upper bound?
– J. Abraham
Nov 17 at 12:05
Can you prove that it is a least upper bound?
– J. Abraham
Nov 17 at 12:05
@J.Abraham All numbers in the set are $le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g.
– Henno Brandsma
Nov 17 at 12:17
@J.Abraham All numbers in the set are $le 1$, and $1$ is assumed for $(a,b,c)=(1,0,0)$, e.g.
– Henno Brandsma
Nov 17 at 12:17
@Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$.
– Sambo
Nov 17 at 12:38
@Henno $(1,0,0)$ is not a valid solution since we require all numbers to be greater than $0$.
– Sambo
Nov 17 at 12:38
1
1
@Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$.
– Henno Brandsma
Nov 17 at 12:40
@Sambo true, use $t,t,1-2t$ for small $t$ instead. We can get as close as we like to $1$.
– Henno Brandsma
Nov 17 at 12:40
add a comment |
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The numbers are positive reals.
– J. Abraham
Nov 17 at 11:52