If $a,b,c$ are positive real numbers and $(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$, then $a+b+c=3$ [duplicate]
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Condition on a,b and c satisfying an equation(TIFR GS 2017)
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Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$
I think this is trivial but I don't know how to prove this fact.
Any Help will be appreciated
real-analysis analysis
edited Nov 29 at 8:45
Asaf Karagila♦
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asked Nov 29 at 5:56
MathLover
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marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, user21820, Henry, José Carlos Santos real-analysis
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This question already has an answer here:
Condition on a,b and c satisfying an equation(TIFR GS 2017)
3 answers
Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$
I think this is trivial but I don't know how to prove this fact.
Any Help will be appreciated
real-analysis analysis
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
asked Nov 29 at 5:56
MathLover
44010
marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, user21820, Henry, José Carlos Santos real-analysis
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Nov 29 at 9:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59
1
Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00
@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05
You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07
Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19
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up vote
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favorite
This question already has an answer here:
Condition on a,b and c satisfying an equation(TIFR GS 2017)
3 answers
Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$
I think this is trivial but I don't know how to prove this fact.
Any Help will be appreciated
real-analysis analysis
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
asked Nov 29 at 5:56
MathLover
44010
This question already has an answer here:
Condition on a,b and c satisfying an equation(TIFR GS 2017)
3 answers
Let $a,b,c$ be positive real number such that
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright)=16$$
then $a+b+c=3$
I think this is trivial but I don't know how to prove this fact.
Any Help will be appreciated
This question already has an answer here:
Condition on a,b and c satisfying an equation(TIFR GS 2017)
3 answers
real-analysis analysis
real-analysis analysis
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
asked Nov 29 at 5:56
MathLover
44010
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
asked Nov 29 at 5:56
MathLover
44010
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
300k32422751
asked Nov 29 at 5:56
MathLover
44010
asked Nov 29 at 5:56
MathLover
44010
asked Nov 29 at 5:56
MathLover
44010
44010
marked as duplicate by Chinnapparaj R, Lord Shark the Unknown, user21820, Henry, José Carlos Santos real-analysis
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Nov 29 at 9:53
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59
1
Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00
@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05
You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07
Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19
|
show 1 more comment
it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59
1
Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00
@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05
You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07
Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19
it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59
it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59
1
1
Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00
Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00
@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05
@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05
You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07
You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07
Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19
Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19
|
show 1 more comment
3 Answers
3
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3
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Use the AM-HM inequality of $1, a, b, c$.
So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$
This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.
The equality becomes an equation only when all four elements are equal.
i.e. $1=a=b=c$. So $a + b + c = 3$.
edited Nov 29 at 6:54
KM101
3,416417
answered Nov 29 at 6:48
Bhargav Kale
412
add a comment |
up vote
3
down vote
Hint:
Using AM HM inequality for non-negative numbers,
$$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
add a comment |
up vote
2
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Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$
answered Nov 29 at 6:48
trancelocation
8,6321520
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Use the AM-HM inequality of $1, a, b, c$.
So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$
This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.
The equality becomes an equation only when all four elements are equal.
i.e. $1=a=b=c$. So $a + b + c = 3$.
edited Nov 29 at 6:54
KM101
3,416417
answered Nov 29 at 6:48
Bhargav Kale
412
add a comment |
up vote
3
down vote
Use the AM-HM inequality of $1, a, b, c$.
So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$
This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.
The equality becomes an equation only when all four elements are equal.
i.e. $1=a=b=c$. So $a + b + c = 3$.
edited Nov 29 at 6:54
KM101
3,416417
answered Nov 29 at 6:48
Bhargav Kale
412
add a comment |
up vote
3
down vote
up vote
3
down vote
Use the AM-HM inequality of $1, a, b, c$.
So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$
This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.
The equality becomes an equation only when all four elements are equal.
i.e. $1=a=b=c$. So $a + b + c = 3$.
edited Nov 29 at 6:54
KM101
3,416417
answered Nov 29 at 6:48
Bhargav Kale
412
Use the AM-HM inequality of $1, a, b, c$.
So $$frac1 + a + b + c4 geq frac41 + frac1a + frac1b + frac1c$$
This gives $$(1 + a + b + c)bigg(1 + frac1a + frac1b + frac1cbigg) geq 16$$.
The equality becomes an equation only when all four elements are equal.
i.e. $1=a=b=c$. So $a + b + c = 3$.
edited Nov 29 at 6:54
KM101
3,416417
answered Nov 29 at 6:48
Bhargav Kale
412
edited Nov 29 at 6:54
KM101
3,416417
edited Nov 29 at 6:54
KM101
3,416417
edited Nov 29 at 6:54
KM101
3,416417
3,416417
answered Nov 29 at 6:48
Bhargav Kale
412
answered Nov 29 at 6:48
Bhargav Kale
412
answered Nov 29 at 6:48
Bhargav Kale
412
412
add a comment |
add a comment |
up vote
3
down vote
Hint:
Using AM HM inequality for non-negative numbers,
$$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
add a comment |
up vote
3
down vote
Hint:
Using AM HM inequality for non-negative numbers,
$$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
add a comment |
up vote
3
down vote
up vote
3
down vote
Hint:
Using AM HM inequality for non-negative numbers,
$$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
Hint:
Using AM HM inequality for non-negative numbers,
$$dfracsum_r=1^na_rngedfrac nsum_r=1^ndfrac1a_r$$
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
edited Nov 29 at 7:04
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
221k15155273
add a comment |
add a comment |
up vote
2
down vote
Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$
answered Nov 29 at 6:48
trancelocation
8,6321520
add a comment |
up vote
2
down vote
Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$
answered Nov 29 at 6:48
trancelocation
8,6321520
add a comment |
up vote
2
down vote
up vote
2
down vote
Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$
answered Nov 29 at 6:48
trancelocation
8,6321520
Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac1a+frac1b+frac1cright) = left(1^2 + sum_cycleft(sqrtaright)^2 right)left(1^2 + sum_cycleft(frac1sqrtaright)^2 right)$$ $$stackrelC.-S.geq(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrta : sqrtb : sqrtc)^T = lambda cdot left( 1: frac1sqrta : : frac1sqrtb : : frac1sqrtcright)^T Rightarrow a=b=c = 1$
answered Nov 29 at 6:48
trancelocation
8,6321520
edited Nov 29 at 6:53
answered Nov 29 at 6:48
trancelocation
8,6321520
answered Nov 29 at 6:48
trancelocation
8,6321520
answered Nov 29 at 6:48
trancelocation
8,6321520
8,6321520
add a comment |
add a comment |
it must be wrong, let b=1, c=1, it is clear that the solution for a is not 1
– MoonKnight
Nov 29 at 5:59
1
Do you mean $(1+a+b+c)(1+1/a+1/b+1/c)=16$?
– Lord Shark the Unknown
Nov 29 at 6:00
@MathLover: I edit your question. I remove the plus sign in the middle. Because if it were then this result is wrong.If you feel I'm missing something, kindly go ahead and edit again!
– Chinnapparaj R
Nov 29 at 6:05
You have edited right ...Thanks a lot
– MathLover
Nov 29 at 6:07
Your question is already appeared on this site. Hold on ....I mention that link!
– Chinnapparaj R
Nov 29 at 6:19