If $a,b,c$ are positive real numbers and $(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right)=16$, then...
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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
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right)=16$$
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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Nov 29 at 9:53
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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
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right)=16$$
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
44210
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
0
down vote
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up vote
0
down vote
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
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right)=16$$
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
44210
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
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right)=16$$
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
44210
edited Nov 29 at 8:45
Asaf Karagila♦
300k32422751
asked Nov 29 at 5:56
MathLover
44210
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
44210
asked Nov 29 at 5:56
MathLover
44210
asked Nov 29 at 5:56
MathLover
44210
44210
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
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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
active
oldest
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up vote
3
down vote
Use the AM-HM inequality of $1, a, b, c$.
So $$frac{1 + a + b + c}{4} geq frac{4}{1 + frac{1}{a} + frac{1}{b} + frac{1}{c}}$$
This gives $$(1 + a + b + c)bigg(1 + frac{1}{a} + frac{1}{b} + frac{1}{c}bigg) geq 16$$.
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,
$$dfrac{sum_{r=1}^na_r}ngedfrac n{sum_{r=1}^ndfrac1{a_r}}$$
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
add a comment |
up vote
2
down vote
Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right) = left(1^2 + sum_{cyc}left(sqrt{a}right)^2 right)left(1^2 + sum_{cyc}left(frac{1}{sqrt{a}}right)^2 right)$$ $$stackrel{C.-S.}{geq}(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrt{a} : sqrt{b} : sqrt{c})^T = lambda cdot left( 1: frac{1}{sqrt{a}} : : frac{1}{sqrt{b}} : : frac{1}{sqrt{c}}right)^T Rightarrow a=b=c = 1$
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 $$frac{1 + a + b + c}{4} geq frac{4}{1 + frac{1}{a} + frac{1}{b} + frac{1}{c}}$$
This gives $$(1 + a + b + c)bigg(1 + frac{1}{a} + frac{1}{b} + frac{1}{c}bigg) geq 16$$.
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 $$frac{1 + a + b + c}{4} geq frac{4}{1 + frac{1}{a} + frac{1}{b} + frac{1}{c}}$$
This gives $$(1 + a + b + c)bigg(1 + frac{1}{a} + frac{1}{b} + frac{1}{c}bigg) geq 16$$.
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 $$frac{1 + a + b + c}{4} geq frac{4}{1 + frac{1}{a} + frac{1}{b} + frac{1}{c}}$$
This gives $$(1 + a + b + c)bigg(1 + frac{1}{a} + frac{1}{b} + frac{1}{c}bigg) geq 16$$.
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 $$frac{1 + a + b + c}{4} geq frac{4}{1 + frac{1}{a} + frac{1}{b} + frac{1}{c}}$$
This gives $$(1 + a + b + c)bigg(1 + frac{1}{a} + frac{1}{b} + frac{1}{c}bigg) geq 16$$.
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,
$$dfrac{sum_{r=1}^na_r}ngedfrac n{sum_{r=1}^ndfrac1{a_r}}$$
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,
$$dfrac{sum_{r=1}^na_r}ngedfrac n{sum_{r=1}^ndfrac1{a_r}}$$
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,
$$dfrac{sum_{r=1}^na_r}ngedfrac n{sum_{r=1}^ndfrac1{a_r}}$$
answered Nov 29 at 5:59
lab bhattacharjee
221k15155273
Hint:
Using AM HM inequality for non-negative numbers,
$$dfrac{sum_{r=1}^na_r}ngedfrac n{sum_{r=1}^ndfrac1{a_r}}$$
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 $a=b=c=1$:
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right) = left(1^2 + sum_{cyc}left(sqrt{a}right)^2 right)left(1^2 + sum_{cyc}left(frac{1}{sqrt{a}}right)^2 right)$$ $$stackrel{C.-S.}{geq}(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrt{a} : sqrt{b} : sqrt{c})^T = lambda cdot left( 1: frac{1}{sqrt{a}} : : frac{1}{sqrt{b}} : : frac{1}{sqrt{c}}right)^T Rightarrow a=b=c = 1$
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 $a=b=c=1$:
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right) = left(1^2 + sum_{cyc}left(sqrt{a}right)^2 right)left(1^2 + sum_{cyc}left(frac{1}{sqrt{a}}right)^2 right)$$ $$stackrel{C.-S.}{geq}(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrt{a} : sqrt{b} : sqrt{c})^T = lambda cdot left( 1: frac{1}{sqrt{a}} : : frac{1}{sqrt{b}} : : frac{1}{sqrt{c}}right)^T Rightarrow a=b=c = 1$
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 $a=b=c=1$:
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right) = left(1^2 + sum_{cyc}left(sqrt{a}right)^2 right)left(1^2 + sum_{cyc}left(frac{1}{sqrt{a}}right)^2 right)$$ $$stackrel{C.-S.}{geq}(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrt{a} : sqrt{b} : sqrt{c})^T = lambda cdot left( 1: frac{1}{sqrt{a}} : : frac{1}{sqrt{b}} : : frac{1}{sqrt{c}}right)^T Rightarrow a=b=c = 1$
answered Nov 29 at 6:48
trancelocation
8,6321520
Using Cauchy-Schwarz inequality you can even show that $a=b=c=1$:
$$(1+a+b+c)left(1+frac{1}{a}+frac{1}{b}+frac{1}{c}right) = left(1^2 + sum_{cyc}left(sqrt{a}right)^2 right)left(1^2 + sum_{cyc}left(frac{1}{sqrt{a}}right)^2 right)$$ $$stackrel{C.-S.}{geq}(1+1+1+1)^2 = 16$$
Equality holds for $(1: sqrt{a} : sqrt{b} : sqrt{c})^T = lambda cdot left( 1: frac{1}{sqrt{a}} : : frac{1}{sqrt{b}} : : frac{1}{sqrt{c}}right)^T Rightarrow a=b=c = 1$
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