If $a, b, c$ are in geometric progression and $a+b+c=abc$ then what is minimum value of $a^4+a^2+7$?
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If $a, b, c$ are in geometric progression and $a+b+c=abc$ then what is minimum value of $a^4+a^2+7$?(a,b,c are positive real numbers)
sequences-and-series
asked Nov 15 at 9:21
Harry Potter
396
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Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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show 3 more comments
up vote
-2
down vote
favorite
If $a, b, c$ are in geometric progression and $a+b+c=abc$ then what is minimum value of $a^4+a^2+7$?(a,b,c are positive real numbers)
sequences-and-series
asked Nov 15 at 9:21
Harry Potter
396
New contributor
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
Hint: Let $b=ar$ and $c=ar^2$.
– Anurag A
Nov 15 at 9:23
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 15 at 9:25
Contest question? What's the source, please?
– Gerry Myerson
Nov 15 at 9:27
Are $a, b, c$ integers?
– Mees de Vries
Nov 15 at 9:39
1
The fact that the solution is trivial (despite your late addition of constraints) makes the problem statement suspicious. Double check it.
– Yves Daoust
Nov 15 at 10:05
|
show 3 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
If $a, b, c$ are in geometric progression and $a+b+c=abc$ then what is minimum value of $a^4+a^2+7$?(a,b,c are positive real numbers)
sequences-and-series
asked Nov 15 at 9:21
Harry Potter
396
New contributor
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If $a, b, c$ are in geometric progression and $a+b+c=abc$ then what is minimum value of $a^4+a^2+7$?(a,b,c are positive real numbers)
sequences-and-series
sequences-and-series
asked Nov 15 at 9:21
Harry Potter
396
New contributor
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 9:21
Harry Potter
396
New contributor
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 15 at 9:48
asked Nov 15 at 9:21
Harry Potter
396
New contributor
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 9:21
Harry Potter
396
asked Nov 15 at 9:21
Harry Potter
396
396
New contributor
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Harry Potter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
Hint: Let $b=ar$ and $c=ar^2$.
– Anurag A
Nov 15 at 9:23
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 15 at 9:25
Contest question? What's the source, please?
– Gerry Myerson
Nov 15 at 9:27
Are $a, b, c$ integers?
– Mees de Vries
Nov 15 at 9:39
1
The fact that the solution is trivial (despite your late addition of constraints) makes the problem statement suspicious. Double check it.
– Yves Daoust
Nov 15 at 10:05
|
show 3 more comments
4
Hint: Let $b=ar$ and $c=ar^2$.
– Anurag A
Nov 15 at 9:23
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 15 at 9:25
Contest question? What's the source, please?
– Gerry Myerson
Nov 15 at 9:27
Are $a, b, c$ integers?
– Mees de Vries
Nov 15 at 9:39
1
The fact that the solution is trivial (despite your late addition of constraints) makes the problem statement suspicious. Double check it.
– Yves Daoust
Nov 15 at 10:05
4
4
Hint: Let $b=ar$ and $c=ar^2$.
– Anurag A
Nov 15 at 9:23
Hint: Let $b=ar$ and $c=ar^2$.
– Anurag A
Nov 15 at 9:23
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 15 at 9:25
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 15 at 9:25
Contest question? What's the source, please?
– Gerry Myerson
Nov 15 at 9:27
Contest question? What's the source, please?
– Gerry Myerson
Nov 15 at 9:27
Are $a, b, c$ integers?
– Mees de Vries
Nov 15 at 9:39
Are $a, b, c$ integers?
– Mees de Vries
Nov 15 at 9:39
1
1
The fact that the solution is trivial (despite your late addition of constraints) makes the problem statement suspicious. Double check it.
– Yves Daoust
Nov 15 at 10:05
The fact that the solution is trivial (despite your late addition of constraints) makes the problem statement suspicious. Double check it.
– Yves Daoust
Nov 15 at 10:05
|
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
Clearly,
$$a^4+a^2+7ge7$$
and equality is achieved by $a=0$ (which is not allowed).
From the given,
$$a(1+r+r^2)=a^3r^3$$
and for $rne0$,
$$a^2=frac{1+r+r^2}{r^3}$$ so that $a$ can be made a small as you want. There is no minimum, just an infimum.
answered Nov 15 at 9:56
Yves Daoust
121k668218
You should treat the case $a = 0$ separately, since you need $ane0$ in order to get $a^2=frac{1+r+r^2}{r^3}.$
– Rchn
Nov 15 at 10:04
@Rchn: right, rewriting. ($a>0$ is given.)
– Yves Daoust
Nov 15 at 10:12
add a comment |
up vote
1
down vote
7
When a=b=c=0
At first I thought (and commented, and was ninja'd) that this was a spurious solution. The reason I'm now calling it an answer is that it's also the infimum of all non-degenerate answers.
Let $b=ar$ and $c=ar^2$
The condition is satisfied iff:
$a + ar + ar^2 = a^3r^3$
This gives a cubic in r: $a^2r^3-r^2-r-1=0$
This cubic is -1 at r=0, and is positive in the limit r->infinity, so it must have a positive root. Thus there exists b and c that satisfy this condition for any a. Therefore a can be chosen to be arbitrarily small.
So you can get as close to 7 as you like with "non-degenerate" answers. So either it's the answer or there is no answer.
answered Nov 15 at 9:55
Steven Irrgang
70126
See question ,a b c are positive and you are saying a=b=c=0
– Harry Potter
Nov 15 at 9:59
Due to Descarte's rule of signs has the equation $a^2r^3-r^2-r-1=0$ exactly one positive root $r.$ However, this root verifies $r>1/a$ from where $b,c>1.$
– user376343
Nov 15 at 10:09
There is no need to care about the roots of the cubic, you can work with $a=f(r)$ rather than $r=f(a)$.
– Yves Daoust
Nov 15 at 10:16
I know it @Yves. I just wanted to point out the wrong argumentation.
– user376343
Nov 15 at 10:26
1
@user376343: I was addressing Steven.
– Yves Daoust
Nov 15 at 11:10
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Clearly,
$$a^4+a^2+7ge7$$
and equality is achieved by $a=0$ (which is not allowed).
From the given,
$$a(1+r+r^2)=a^3r^3$$
and for $rne0$,
$$a^2=frac{1+r+r^2}{r^3}$$ so that $a$ can be made a small as you want. There is no minimum, just an infimum.
answered Nov 15 at 9:56
Yves Daoust
121k668218
You should treat the case $a = 0$ separately, since you need $ane0$ in order to get $a^2=frac{1+r+r^2}{r^3}.$
– Rchn
Nov 15 at 10:04
@Rchn: right, rewriting. ($a>0$ is given.)
– Yves Daoust
Nov 15 at 10:12
add a comment |
up vote
4
down vote
Clearly,
$$a^4+a^2+7ge7$$
and equality is achieved by $a=0$ (which is not allowed).
From the given,
$$a(1+r+r^2)=a^3r^3$$
and for $rne0$,
$$a^2=frac{1+r+r^2}{r^3}$$ so that $a$ can be made a small as you want. There is no minimum, just an infimum.
answered Nov 15 at 9:56
Yves Daoust
121k668218
You should treat the case $a = 0$ separately, since you need $ane0$ in order to get $a^2=frac{1+r+r^2}{r^3}.$
– Rchn
Nov 15 at 10:04
@Rchn: right, rewriting. ($a>0$ is given.)
– Yves Daoust
Nov 15 at 10:12
add a comment |
up vote
4
down vote
up vote
4
down vote
Clearly,
$$a^4+a^2+7ge7$$
and equality is achieved by $a=0$ (which is not allowed).
From the given,
$$a(1+r+r^2)=a^3r^3$$
and for $rne0$,
$$a^2=frac{1+r+r^2}{r^3}$$ so that $a$ can be made a small as you want. There is no minimum, just an infimum.
answered Nov 15 at 9:56
Yves Daoust
121k668218
Clearly,
$$a^4+a^2+7ge7$$
and equality is achieved by $a=0$ (which is not allowed).
From the given,
$$a(1+r+r^2)=a^3r^3$$
and for $rne0$,
$$a^2=frac{1+r+r^2}{r^3}$$ so that $a$ can be made a small as you want. There is no minimum, just an infimum.
answered Nov 15 at 9:56
Yves Daoust
121k668218
edited Nov 15 at 10:07
answered Nov 15 at 9:56
Yves Daoust
121k668218
answered Nov 15 at 9:56
Yves Daoust
121k668218
answered Nov 15 at 9:56
Yves Daoust
121k668218
121k668218
You should treat the case $a = 0$ separately, since you need $ane0$ in order to get $a^2=frac{1+r+r^2}{r^3}.$
– Rchn
Nov 15 at 10:04
@Rchn: right, rewriting. ($a>0$ is given.)
– Yves Daoust
Nov 15 at 10:12
add a comment |
You should treat the case $a = 0$ separately, since you need $ane0$ in order to get $a^2=frac{1+r+r^2}{r^3}.$
– Rchn
Nov 15 at 10:04
@Rchn: right, rewriting. ($a>0$ is given.)
– Yves Daoust
Nov 15 at 10:12
You should treat the case $a = 0$ separately, since you need $ane0$ in order to get $a^2=frac{1+r+r^2}{r^3}.$
– Rchn
Nov 15 at 10:04
You should treat the case $a = 0$ separately, since you need $ane0$ in order to get $a^2=frac{1+r+r^2}{r^3}.$
– Rchn
Nov 15 at 10:04
@Rchn: right, rewriting. ($a>0$ is given.)
– Yves Daoust
Nov 15 at 10:12
@Rchn: right, rewriting. ($a>0$ is given.)
– Yves Daoust
Nov 15 at 10:12
add a comment |
up vote
1
down vote
7
When a=b=c=0
At first I thought (and commented, and was ninja'd) that this was a spurious solution. The reason I'm now calling it an answer is that it's also the infimum of all non-degenerate answers.
Let $b=ar$ and $c=ar^2$
The condition is satisfied iff:
$a + ar + ar^2 = a^3r^3$
This gives a cubic in r: $a^2r^3-r^2-r-1=0$
This cubic is -1 at r=0, and is positive in the limit r->infinity, so it must have a positive root. Thus there exists b and c that satisfy this condition for any a. Therefore a can be chosen to be arbitrarily small.
So you can get as close to 7 as you like with "non-degenerate" answers. So either it's the answer or there is no answer.
answered Nov 15 at 9:55
Steven Irrgang
70126
See question ,a b c are positive and you are saying a=b=c=0
– Harry Potter
Nov 15 at 9:59
Due to Descarte's rule of signs has the equation $a^2r^3-r^2-r-1=0$ exactly one positive root $r.$ However, this root verifies $r>1/a$ from where $b,c>1.$
– user376343
Nov 15 at 10:09
There is no need to care about the roots of the cubic, you can work with $a=f(r)$ rather than $r=f(a)$.
– Yves Daoust
Nov 15 at 10:16
I know it @Yves. I just wanted to point out the wrong argumentation.
– user376343
Nov 15 at 10:26
1
@user376343: I was addressing Steven.
– Yves Daoust
Nov 15 at 11:10
add a comment |
up vote
1
down vote
7
When a=b=c=0
At first I thought (and commented, and was ninja'd) that this was a spurious solution. The reason I'm now calling it an answer is that it's also the infimum of all non-degenerate answers.
Let $b=ar$ and $c=ar^2$
The condition is satisfied iff:
$a + ar + ar^2 = a^3r^3$
This gives a cubic in r: $a^2r^3-r^2-r-1=0$
This cubic is -1 at r=0, and is positive in the limit r->infinity, so it must have a positive root. Thus there exists b and c that satisfy this condition for any a. Therefore a can be chosen to be arbitrarily small.
So you can get as close to 7 as you like with "non-degenerate" answers. So either it's the answer or there is no answer.
answered Nov 15 at 9:55
Steven Irrgang
70126
See question ,a b c are positive and you are saying a=b=c=0
– Harry Potter
Nov 15 at 9:59
Due to Descarte's rule of signs has the equation $a^2r^3-r^2-r-1=0$ exactly one positive root $r.$ However, this root verifies $r>1/a$ from where $b,c>1.$
– user376343
Nov 15 at 10:09
There is no need to care about the roots of the cubic, you can work with $a=f(r)$ rather than $r=f(a)$.
– Yves Daoust
Nov 15 at 10:16
I know it @Yves. I just wanted to point out the wrong argumentation.
– user376343
Nov 15 at 10:26
1
@user376343: I was addressing Steven.
– Yves Daoust
Nov 15 at 11:10
add a comment |
up vote
1
down vote
up vote
1
down vote
7
When a=b=c=0
At first I thought (and commented, and was ninja'd) that this was a spurious solution. The reason I'm now calling it an answer is that it's also the infimum of all non-degenerate answers.
Let $b=ar$ and $c=ar^2$
The condition is satisfied iff:
$a + ar + ar^2 = a^3r^3$
This gives a cubic in r: $a^2r^3-r^2-r-1=0$
This cubic is -1 at r=0, and is positive in the limit r->infinity, so it must have a positive root. Thus there exists b and c that satisfy this condition for any a. Therefore a can be chosen to be arbitrarily small.
So you can get as close to 7 as you like with "non-degenerate" answers. So either it's the answer or there is no answer.
answered Nov 15 at 9:55
Steven Irrgang
70126
7
When a=b=c=0
At first I thought (and commented, and was ninja'd) that this was a spurious solution. The reason I'm now calling it an answer is that it's also the infimum of all non-degenerate answers.
Let $b=ar$ and $c=ar^2$
The condition is satisfied iff:
$a + ar + ar^2 = a^3r^3$
This gives a cubic in r: $a^2r^3-r^2-r-1=0$
This cubic is -1 at r=0, and is positive in the limit r->infinity, so it must have a positive root. Thus there exists b and c that satisfy this condition for any a. Therefore a can be chosen to be arbitrarily small.
So you can get as close to 7 as you like with "non-degenerate" answers. So either it's the answer or there is no answer.
answered Nov 15 at 9:55
Steven Irrgang
70126
answered Nov 15 at 9:55
Steven Irrgang
70126
answered Nov 15 at 9:55
Steven Irrgang
70126
answered Nov 15 at 9:55
Steven Irrgang
70126
70126
See question ,a b c are positive and you are saying a=b=c=0
– Harry Potter
Nov 15 at 9:59
Due to Descarte's rule of signs has the equation $a^2r^3-r^2-r-1=0$ exactly one positive root $r.$ However, this root verifies $r>1/a$ from where $b,c>1.$
– user376343
Nov 15 at 10:09
There is no need to care about the roots of the cubic, you can work with $a=f(r)$ rather than $r=f(a)$.
– Yves Daoust
Nov 15 at 10:16
I know it @Yves. I just wanted to point out the wrong argumentation.
– user376343
Nov 15 at 10:26
1
@user376343: I was addressing Steven.
– Yves Daoust
Nov 15 at 11:10
add a comment |
See question ,a b c are positive and you are saying a=b=c=0
– Harry Potter
Nov 15 at 9:59
Due to Descarte's rule of signs has the equation $a^2r^3-r^2-r-1=0$ exactly one positive root $r.$ However, this root verifies $r>1/a$ from where $b,c>1.$
– user376343
Nov 15 at 10:09
There is no need to care about the roots of the cubic, you can work with $a=f(r)$ rather than $r=f(a)$.
– Yves Daoust
Nov 15 at 10:16
I know it @Yves. I just wanted to point out the wrong argumentation.
– user376343
Nov 15 at 10:26
1
@user376343: I was addressing Steven.
– Yves Daoust
Nov 15 at 11:10
See question ,a b c are positive and you are saying a=b=c=0
– Harry Potter
Nov 15 at 9:59
See question ,a b c are positive and you are saying a=b=c=0
– Harry Potter
Nov 15 at 9:59
Due to Descarte's rule of signs has the equation $a^2r^3-r^2-r-1=0$ exactly one positive root $r.$ However, this root verifies $r>1/a$ from where $b,c>1.$
– user376343
Nov 15 at 10:09
Due to Descarte's rule of signs has the equation $a^2r^3-r^2-r-1=0$ exactly one positive root $r.$ However, this root verifies $r>1/a$ from where $b,c>1.$
– user376343
Nov 15 at 10:09
There is no need to care about the roots of the cubic, you can work with $a=f(r)$ rather than $r=f(a)$.
– Yves Daoust
Nov 15 at 10:16
There is no need to care about the roots of the cubic, you can work with $a=f(r)$ rather than $r=f(a)$.
– Yves Daoust
Nov 15 at 10:16
I know it @Yves. I just wanted to point out the wrong argumentation.
– user376343
Nov 15 at 10:26
I know it @Yves. I just wanted to point out the wrong argumentation.
– user376343
Nov 15 at 10:26
1
1
@user376343: I was addressing Steven.
– Yves Daoust
Nov 15 at 11:10
@user376343: I was addressing Steven.
– Yves Daoust
Nov 15 at 11:10
add a comment |
Harry Potter is a new contributor. Be nice, and check out our Code of Conduct.
Harry Potter is a new contributor. Be nice, and check out our Code of Conduct.
Harry Potter is a new contributor. Be nice, and check out our Code of Conduct.
Harry Potter is a new contributor. Be nice, and check out our Code of Conduct.
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4
Hint: Let $b=ar$ and $c=ar^2$.
– Anurag A
Nov 15 at 9:23
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Nov 15 at 9:25
Contest question? What's the source, please?
– Gerry Myerson
Nov 15 at 9:27
Are $a, b, c$ integers?
– Mees de Vries
Nov 15 at 9:39
1
The fact that the solution is trivial (despite your late addition of constraints) makes the problem statement suspicious. Double check it.
– Yves Daoust
Nov 15 at 10:05