Proving a three variables inequality
up vote
6
down vote
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Given that $frac{1}{a}+frac{1}{b}+frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(frac{1}{a}+frac{1}{b}+frac{1}{c})$ getting that $a+b+cge9$.
Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+cge3sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)ge63$$The last thing I tought about was that I have both $a+b+cge9$ and $a+b+cge3sqrt[3]{abc}$ so if I somehow related them I would have $sqrt[3]{abc} ge 3 rightarrow abcge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...
I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!
inequality tangent-line-method
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
asked Dec 4 at 20:58
Spasoje Durovic
1548
|
show 1 more comment
up vote
6
down vote
favorite
Given that $frac{1}{a}+frac{1}{b}+frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(frac{1}{a}+frac{1}{b}+frac{1}{c})$ getting that $a+b+cge9$.
Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+cge3sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)ge63$$The last thing I tought about was that I have both $a+b+cge9$ and $a+b+cge3sqrt[3]{abc}$ so if I somehow related them I would have $sqrt[3]{abc} ge 3 rightarrow abcge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...
I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!
inequality tangent-line-method
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
asked Dec 4 at 20:58
Spasoje Durovic
1548
I haven't tried this I admit .... but my instant thought is, since we 'know' (please note the apostrophæ!) that it attains its minimum value at a=b=c=3, could we not hypothesise that it's 3, plug in 3+δ, 3+ε, & 3+ζ, & show that any increase in some measure of the size of δ, ε & ζ causes an increase in the product under the constraint? As I said - just my instantaneous thought on seeing the question!
– AmbretteOrrisey
Dec 4 at 22:54
@AmbretteOrrisey One could take the second partial derivative (since the expression is symetric you can take it with respect to any variable) and see that the function $f(a,b,c)=(a+1)(b+1)(c+1)$ is convex since it's second derivative is equal to zero, given this we basically done what you suggested to (i.e. any increment in a,b,c gives us an increment in f(a,b,c)) but the thing is, you'd still have to formally prove that a=b=c=3 it's the minimal value of the funcion which really doesn't seem easy to me (in one of the answers they actually proved that 3 is the minimal value, you can look it up)
– Spasoje Durovic
Dec 5 at 20:05
@ Spasoje Durovic -- So you can without too much difficulty prove the way said maybe you could? ... assuming that the minimum is at a=b=c=3 ... but then the big stumbilngblock is proving that the minimum is at that? It's funny how so often the stumbilngblock of a proof is the seemingly most trivial part of it!
– AmbretteOrrisey
Dec 5 at 22:03
But do you really need to prove that, logically to prove this theorem as it stands? ... as you can just calculate that at a=b=c=3 the product in the theorem is 64 -- so if you prove that any departure (satisfying the constraint) from that point can only increase that product, surely then you've proven the theorem as it stands? Although I see that if you only have convexity, then you would also need to prove that the minimum is at that particular point.
– AmbretteOrrisey
Dec 5 at 22:17
1
And it looks promising on the face of it! Maybe what you've said then explains why none of the people here have done it that way!
– AmbretteOrrisey
Dec 7 at 15:40
|
show 1 more comment
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Given that $frac{1}{a}+frac{1}{b}+frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(frac{1}{a}+frac{1}{b}+frac{1}{c})$ getting that $a+b+cge9$.
Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+cge3sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)ge63$$The last thing I tought about was that I have both $a+b+cge9$ and $a+b+cge3sqrt[3]{abc}$ so if I somehow related them I would have $sqrt[3]{abc} ge 3 rightarrow abcge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...
I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!
inequality tangent-line-method
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
asked Dec 4 at 20:58
Spasoje Durovic
1548
Given that $frac{1}{a}+frac{1}{b}+frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(frac{1}{a}+frac{1}{b}+frac{1}{c})$ getting that $a+b+cge9$.
Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+cge3sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)ge63$$The last thing I tought about was that I have both $a+b+cge9$ and $a+b+cge3sqrt[3]{abc}$ so if I somehow related them I would have $sqrt[3]{abc} ge 3 rightarrow abcge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...
I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!
inequality tangent-line-method
inequality tangent-line-method
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
asked Dec 4 at 20:58
Spasoje Durovic
1548
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
asked Dec 4 at 20:58
Spasoje Durovic
1548
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
edited Dec 5 at 5:33
Michael Rozenberg
94.8k1588183
94.8k1588183
asked Dec 4 at 20:58
Spasoje Durovic
1548
asked Dec 4 at 20:58
Spasoje Durovic
1548
asked Dec 4 at 20:58
Spasoje Durovic
1548
1548
I haven't tried this I admit .... but my instant thought is, since we 'know' (please note the apostrophæ!) that it attains its minimum value at a=b=c=3, could we not hypothesise that it's 3, plug in 3+δ, 3+ε, & 3+ζ, & show that any increase in some measure of the size of δ, ε & ζ causes an increase in the product under the constraint? As I said - just my instantaneous thought on seeing the question!
– AmbretteOrrisey
Dec 4 at 22:54
@AmbretteOrrisey One could take the second partial derivative (since the expression is symetric you can take it with respect to any variable) and see that the function $f(a,b,c)=(a+1)(b+1)(c+1)$ is convex since it's second derivative is equal to zero, given this we basically done what you suggested to (i.e. any increment in a,b,c gives us an increment in f(a,b,c)) but the thing is, you'd still have to formally prove that a=b=c=3 it's the minimal value of the funcion which really doesn't seem easy to me (in one of the answers they actually proved that 3 is the minimal value, you can look it up)
– Spasoje Durovic
Dec 5 at 20:05
@ Spasoje Durovic -- So you can without too much difficulty prove the way said maybe you could? ... assuming that the minimum is at a=b=c=3 ... but then the big stumbilngblock is proving that the minimum is at that? It's funny how so often the stumbilngblock of a proof is the seemingly most trivial part of it!
– AmbretteOrrisey
Dec 5 at 22:03
But do you really need to prove that, logically to prove this theorem as it stands? ... as you can just calculate that at a=b=c=3 the product in the theorem is 64 -- so if you prove that any departure (satisfying the constraint) from that point can only increase that product, surely then you've proven the theorem as it stands? Although I see that if you only have convexity, then you would also need to prove that the minimum is at that particular point.
– AmbretteOrrisey
Dec 5 at 22:17
1
And it looks promising on the face of it! Maybe what you've said then explains why none of the people here have done it that way!
– AmbretteOrrisey
Dec 7 at 15:40
|
show 1 more comment
I haven't tried this I admit .... but my instant thought is, since we 'know' (please note the apostrophæ!) that it attains its minimum value at a=b=c=3, could we not hypothesise that it's 3, plug in 3+δ, 3+ε, & 3+ζ, & show that any increase in some measure of the size of δ, ε & ζ causes an increase in the product under the constraint? As I said - just my instantaneous thought on seeing the question!
– AmbretteOrrisey
Dec 4 at 22:54
@AmbretteOrrisey One could take the second partial derivative (since the expression is symetric you can take it with respect to any variable) and see that the function $f(a,b,c)=(a+1)(b+1)(c+1)$ is convex since it's second derivative is equal to zero, given this we basically done what you suggested to (i.e. any increment in a,b,c gives us an increment in f(a,b,c)) but the thing is, you'd still have to formally prove that a=b=c=3 it's the minimal value of the funcion which really doesn't seem easy to me (in one of the answers they actually proved that 3 is the minimal value, you can look it up)
– Spasoje Durovic
Dec 5 at 20:05
@ Spasoje Durovic -- So you can without too much difficulty prove the way said maybe you could? ... assuming that the minimum is at a=b=c=3 ... but then the big stumbilngblock is proving that the minimum is at that? It's funny how so often the stumbilngblock of a proof is the seemingly most trivial part of it!
– AmbretteOrrisey
Dec 5 at 22:03
But do you really need to prove that, logically to prove this theorem as it stands? ... as you can just calculate that at a=b=c=3 the product in the theorem is 64 -- so if you prove that any departure (satisfying the constraint) from that point can only increase that product, surely then you've proven the theorem as it stands? Although I see that if you only have convexity, then you would also need to prove that the minimum is at that particular point.
– AmbretteOrrisey
Dec 5 at 22:17
1
And it looks promising on the face of it! Maybe what you've said then explains why none of the people here have done it that way!
– AmbretteOrrisey
Dec 7 at 15:40
I haven't tried this I admit .... but my instant thought is, since we 'know' (please note the apostrophæ!) that it attains its minimum value at a=b=c=3, could we not hypothesise that it's 3, plug in 3+δ, 3+ε, & 3+ζ, & show that any increase in some measure of the size of δ, ε & ζ causes an increase in the product under the constraint? As I said - just my instantaneous thought on seeing the question!
– AmbretteOrrisey
Dec 4 at 22:54
I haven't tried this I admit .... but my instant thought is, since we 'know' (please note the apostrophæ!) that it attains its minimum value at a=b=c=3, could we not hypothesise that it's 3, plug in 3+δ, 3+ε, & 3+ζ, & show that any increase in some measure of the size of δ, ε & ζ causes an increase in the product under the constraint? As I said - just my instantaneous thought on seeing the question!
– AmbretteOrrisey
Dec 4 at 22:54
@AmbretteOrrisey One could take the second partial derivative (since the expression is symetric you can take it with respect to any variable) and see that the function $f(a,b,c)=(a+1)(b+1)(c+1)$ is convex since it's second derivative is equal to zero, given this we basically done what you suggested to (i.e. any increment in a,b,c gives us an increment in f(a,b,c)) but the thing is, you'd still have to formally prove that a=b=c=3 it's the minimal value of the funcion which really doesn't seem easy to me (in one of the answers they actually proved that 3 is the minimal value, you can look it up)
– Spasoje Durovic
Dec 5 at 20:05
@AmbretteOrrisey One could take the second partial derivative (since the expression is symetric you can take it with respect to any variable) and see that the function $f(a,b,c)=(a+1)(b+1)(c+1)$ is convex since it's second derivative is equal to zero, given this we basically done what you suggested to (i.e. any increment in a,b,c gives us an increment in f(a,b,c)) but the thing is, you'd still have to formally prove that a=b=c=3 it's the minimal value of the funcion which really doesn't seem easy to me (in one of the answers they actually proved that 3 is the minimal value, you can look it up)
– Spasoje Durovic
Dec 5 at 20:05
@ Spasoje Durovic -- So you can without too much difficulty prove the way said maybe you could? ... assuming that the minimum is at a=b=c=3 ... but then the big stumbilngblock is proving that the minimum is at that? It's funny how so often the stumbilngblock of a proof is the seemingly most trivial part of it!
– AmbretteOrrisey
Dec 5 at 22:03
@ Spasoje Durovic -- So you can without too much difficulty prove the way said maybe you could? ... assuming that the minimum is at a=b=c=3 ... but then the big stumbilngblock is proving that the minimum is at that? It's funny how so often the stumbilngblock of a proof is the seemingly most trivial part of it!
– AmbretteOrrisey
Dec 5 at 22:03
But do you really need to prove that, logically to prove this theorem as it stands? ... as you can just calculate that at a=b=c=3 the product in the theorem is 64 -- so if you prove that any departure (satisfying the constraint) from that point can only increase that product, surely then you've proven the theorem as it stands? Although I see that if you only have convexity, then you would also need to prove that the minimum is at that particular point.
– AmbretteOrrisey
Dec 5 at 22:17
But do you really need to prove that, logically to prove this theorem as it stands? ... as you can just calculate that at a=b=c=3 the product in the theorem is 64 -- so if you prove that any departure (satisfying the constraint) from that point can only increase that product, surely then you've proven the theorem as it stands? Although I see that if you only have convexity, then you would also need to prove that the minimum is at that particular point.
– AmbretteOrrisey
Dec 5 at 22:17
1
1
And it looks promising on the face of it! Maybe what you've said then explains why none of the people here have done it that way!
– AmbretteOrrisey
Dec 7 at 15:40
And it looks promising on the face of it! Maybe what you've said then explains why none of the people here have done it that way!
– AmbretteOrrisey
Dec 7 at 15:40
|
show 1 more comment
4 Answers
4
active
oldest
votes
up vote
8
down vote
accepted
By the super-additivity of the geometric mean / convexity of $log(e^x+1)$
$$ (1+a)(1+b)(1+c) geq (1+sqrt[3]{abc})^3 $$
and by the GM-HM inequality
$$ sqrt[3]{abc} geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}=3.$$
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
1
Please explain the downvote. I do not see anything wrong with this approach, it is pretty straightforward.
– Jack D'Aurizio
Dec 4 at 21:05
3
Maybe the downvoter's arithmetic is so bad that they want you explain that $4^3 = 64$ $ddot{smile}$.
– Rob Arthan
Dec 4 at 21:08
1
@JackD'Aurizio I'm convinced that there are users who stays here on MSE only to downvote, probably to give vent to frustration in the real world!
– gimusi
Dec 4 at 21:15
1
@greedoid Nothing personal with the comment, I was only joking of course.
– gimusi
Dec 4 at 21:16
1
@RobArthan Not a theory but only a joke! Bye :)
– gimusi
Dec 4 at 21:20
|
show 8 more comments
up vote
9
down vote
I think you mean $$a,b,c>0$$ in this case we have
$$frac{a+b+c}{3}geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}$$ so $$a+b+cgeq 9$$ and we get also $$frac{bc+ac+ab}{3}geq sqrt[3]{(abc)^2}$$ so $$abcgeq 27$$ and $$ab+ac+bcgeq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1geq 27+27+10=64$$
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
2
Welcome back Dr. Graubner!
– gimusi
Dec 4 at 21:19
Yeah, OP forgot to mention that $a,b,c$ have to be positive, I forgot to add that I had noticed that
– Spasoje Durovic
Dec 4 at 21:20
2
Thanks, i'm also glad to see you!
– Dr. Sonnhard Graubner
Dec 4 at 21:20
add a comment |
up vote
4
down vote
Since $ab+bc+ca= abc$ and $$ab+bc+cageq 3sqrt[3]{a^2b^2c^2}implies abcgeq 27$$
Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1geq 55+3sqrt[3]{27} = 64$$
answered Dec 4 at 21:13
greedoid
36.4k114591
Yup, that first passage was the one I didn't see, I'm actually terrible at AM-GM!...
– Spasoje Durovic
Dec 4 at 21:22
Yeah, I've started working on inequalities about a week ago, hope I'll get better soon! cheers (nice to see you're from the Balkans too :P)
– Spasoje Durovic
Dec 4 at 21:27
Actualy I'm not.
– greedoid
Dec 4 at 21:28
My bad! I supposed it since you used Cyrillic
– Spasoje Durovic
Dec 4 at 21:29
1
Never mind, it is OK
– greedoid
Dec 4 at 21:30
add a comment |
up vote
2
down vote
For positive variables we need to prove that
$$lnleft((1+a)(1+b)(1+c)right)geqln64$$ or
$$sum_{cyc}(ln(1+a)-2ln2)geq0$$ or
$$sum_{cyc}left(ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)right)geq0,$$
which is true because
$$f(a)=ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)geq0$$ for all $a>0$.
Indeed, $$f'(a)=frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
Can you elaborate on how you got the first expression?
– YiFan
Dec 4 at 22:53
1
@YiFan presumably by taking the $ln(bullet)$ of both sides using the fact that $ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $ln(64)$ term into three equal parts to get it into the cyclic sum, those being $ln(64)/3 = ln(4) = 2 ln(2).$
– CR Drost
Dec 4 at 23:55
@CRDrost Of course. Thanks for the help!
– YiFan
Dec 4 at 23:56
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
By the super-additivity of the geometric mean / convexity of $log(e^x+1)$
$$ (1+a)(1+b)(1+c) geq (1+sqrt[3]{abc})^3 $$
and by the GM-HM inequality
$$ sqrt[3]{abc} geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}=3.$$
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
1
Please explain the downvote. I do not see anything wrong with this approach, it is pretty straightforward.
– Jack D'Aurizio
Dec 4 at 21:05
3
Maybe the downvoter's arithmetic is so bad that they want you explain that $4^3 = 64$ $ddot{smile}$.
– Rob Arthan
Dec 4 at 21:08
1
@JackD'Aurizio I'm convinced that there are users who stays here on MSE only to downvote, probably to give vent to frustration in the real world!
– gimusi
Dec 4 at 21:15
1
@greedoid Nothing personal with the comment, I was only joking of course.
– gimusi
Dec 4 at 21:16
1
@RobArthan Not a theory but only a joke! Bye :)
– gimusi
Dec 4 at 21:20
|
show 8 more comments
up vote
8
down vote
accepted
By the super-additivity of the geometric mean / convexity of $log(e^x+1)$
$$ (1+a)(1+b)(1+c) geq (1+sqrt[3]{abc})^3 $$
and by the GM-HM inequality
$$ sqrt[3]{abc} geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}=3.$$
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
1
Please explain the downvote. I do not see anything wrong with this approach, it is pretty straightforward.
– Jack D'Aurizio
Dec 4 at 21:05
3
Maybe the downvoter's arithmetic is so bad that they want you explain that $4^3 = 64$ $ddot{smile}$.
– Rob Arthan
Dec 4 at 21:08
1
@JackD'Aurizio I'm convinced that there are users who stays here on MSE only to downvote, probably to give vent to frustration in the real world!
– gimusi
Dec 4 at 21:15
1
@greedoid Nothing personal with the comment, I was only joking of course.
– gimusi
Dec 4 at 21:16
1
@RobArthan Not a theory but only a joke! Bye :)
– gimusi
Dec 4 at 21:20
|
show 8 more comments
up vote
8
down vote
accepted
up vote
8
down vote
accepted
By the super-additivity of the geometric mean / convexity of $log(e^x+1)$
$$ (1+a)(1+b)(1+c) geq (1+sqrt[3]{abc})^3 $$
and by the GM-HM inequality
$$ sqrt[3]{abc} geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}=3.$$
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
By the super-additivity of the geometric mean / convexity of $log(e^x+1)$
$$ (1+a)(1+b)(1+c) geq (1+sqrt[3]{abc})^3 $$
and by the GM-HM inequality
$$ sqrt[3]{abc} geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}=3.$$
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
answered Dec 4 at 21:04
Jack D'Aurizio
284k33275654
284k33275654
1
Please explain the downvote. I do not see anything wrong with this approach, it is pretty straightforward.
– Jack D'Aurizio
Dec 4 at 21:05
3
Maybe the downvoter's arithmetic is so bad that they want you explain that $4^3 = 64$ $ddot{smile}$.
– Rob Arthan
Dec 4 at 21:08
1
@JackD'Aurizio I'm convinced that there are users who stays here on MSE only to downvote, probably to give vent to frustration in the real world!
– gimusi
Dec 4 at 21:15
1
@greedoid Nothing personal with the comment, I was only joking of course.
– gimusi
Dec 4 at 21:16
1
@RobArthan Not a theory but only a joke! Bye :)
– gimusi
Dec 4 at 21:20
|
show 8 more comments
1
Please explain the downvote. I do not see anything wrong with this approach, it is pretty straightforward.
– Jack D'Aurizio
Dec 4 at 21:05
3
Maybe the downvoter's arithmetic is so bad that they want you explain that $4^3 = 64$ $ddot{smile}$.
– Rob Arthan
Dec 4 at 21:08
1
@JackD'Aurizio I'm convinced that there are users who stays here on MSE only to downvote, probably to give vent to frustration in the real world!
– gimusi
Dec 4 at 21:15
1
@greedoid Nothing personal with the comment, I was only joking of course.
– gimusi
Dec 4 at 21:16
1
@RobArthan Not a theory but only a joke! Bye :)
– gimusi
Dec 4 at 21:20
1
1
Please explain the downvote. I do not see anything wrong with this approach, it is pretty straightforward.
– Jack D'Aurizio
Dec 4 at 21:05
Please explain the downvote. I do not see anything wrong with this approach, it is pretty straightforward.
– Jack D'Aurizio
Dec 4 at 21:05
3
3
Maybe the downvoter's arithmetic is so bad that they want you explain that $4^3 = 64$ $ddot{smile}$.
– Rob Arthan
Dec 4 at 21:08
Maybe the downvoter's arithmetic is so bad that they want you explain that $4^3 = 64$ $ddot{smile}$.
– Rob Arthan
Dec 4 at 21:08
1
1
@JackD'Aurizio I'm convinced that there are users who stays here on MSE only to downvote, probably to give vent to frustration in the real world!
– gimusi
Dec 4 at 21:15
@JackD'Aurizio I'm convinced that there are users who stays here on MSE only to downvote, probably to give vent to frustration in the real world!
– gimusi
Dec 4 at 21:15
1
1
@greedoid Nothing personal with the comment, I was only joking of course.
– gimusi
Dec 4 at 21:16
@greedoid Nothing personal with the comment, I was only joking of course.
– gimusi
Dec 4 at 21:16
1
1
@RobArthan Not a theory but only a joke! Bye :)
– gimusi
Dec 4 at 21:20
@RobArthan Not a theory but only a joke! Bye :)
– gimusi
Dec 4 at 21:20
|
show 8 more comments
up vote
9
down vote
I think you mean $$a,b,c>0$$ in this case we have
$$frac{a+b+c}{3}geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}$$ so $$a+b+cgeq 9$$ and we get also $$frac{bc+ac+ab}{3}geq sqrt[3]{(abc)^2}$$ so $$abcgeq 27$$ and $$ab+ac+bcgeq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1geq 27+27+10=64$$
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
2
Welcome back Dr. Graubner!
– gimusi
Dec 4 at 21:19
Yeah, OP forgot to mention that $a,b,c$ have to be positive, I forgot to add that I had noticed that
– Spasoje Durovic
Dec 4 at 21:20
2
Thanks, i'm also glad to see you!
– Dr. Sonnhard Graubner
Dec 4 at 21:20
add a comment |
up vote
9
down vote
I think you mean $$a,b,c>0$$ in this case we have
$$frac{a+b+c}{3}geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}$$ so $$a+b+cgeq 9$$ and we get also $$frac{bc+ac+ab}{3}geq sqrt[3]{(abc)^2}$$ so $$abcgeq 27$$ and $$ab+ac+bcgeq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1geq 27+27+10=64$$
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
2
Welcome back Dr. Graubner!
– gimusi
Dec 4 at 21:19
Yeah, OP forgot to mention that $a,b,c$ have to be positive, I forgot to add that I had noticed that
– Spasoje Durovic
Dec 4 at 21:20
2
Thanks, i'm also glad to see you!
– Dr. Sonnhard Graubner
Dec 4 at 21:20
add a comment |
up vote
9
down vote
up vote
9
down vote
I think you mean $$a,b,c>0$$ in this case we have
$$frac{a+b+c}{3}geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}$$ so $$a+b+cgeq 9$$ and we get also $$frac{bc+ac+ab}{3}geq sqrt[3]{(abc)^2}$$ so $$abcgeq 27$$ and $$ab+ac+bcgeq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1geq 27+27+10=64$$
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
I think you mean $$a,b,c>0$$ in this case we have
$$frac{a+b+c}{3}geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}$$ so $$a+b+cgeq 9$$ and we get also $$frac{bc+ac+ab}{3}geq sqrt[3]{(abc)^2}$$ so $$abcgeq 27$$ and $$ab+ac+bcgeq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1geq 27+27+10=64$$
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
answered Dec 4 at 21:14
Dr. Sonnhard Graubner
71.9k32864
71.9k32864
2
Welcome back Dr. Graubner!
– gimusi
Dec 4 at 21:19
Yeah, OP forgot to mention that $a,b,c$ have to be positive, I forgot to add that I had noticed that
– Spasoje Durovic
Dec 4 at 21:20
2
Thanks, i'm also glad to see you!
– Dr. Sonnhard Graubner
Dec 4 at 21:20
add a comment |
2
Welcome back Dr. Graubner!
– gimusi
Dec 4 at 21:19
Yeah, OP forgot to mention that $a,b,c$ have to be positive, I forgot to add that I had noticed that
– Spasoje Durovic
Dec 4 at 21:20
2
Thanks, i'm also glad to see you!
– Dr. Sonnhard Graubner
Dec 4 at 21:20
2
2
Welcome back Dr. Graubner!
– gimusi
Dec 4 at 21:19
Welcome back Dr. Graubner!
– gimusi
Dec 4 at 21:19
Yeah, OP forgot to mention that $a,b,c$ have to be positive, I forgot to add that I had noticed that
– Spasoje Durovic
Dec 4 at 21:20
Yeah, OP forgot to mention that $a,b,c$ have to be positive, I forgot to add that I had noticed that
– Spasoje Durovic
Dec 4 at 21:20
2
2
Thanks, i'm also glad to see you!
– Dr. Sonnhard Graubner
Dec 4 at 21:20
Thanks, i'm also glad to see you!
– Dr. Sonnhard Graubner
Dec 4 at 21:20
add a comment |
up vote
4
down vote
Since $ab+bc+ca= abc$ and $$ab+bc+cageq 3sqrt[3]{a^2b^2c^2}implies abcgeq 27$$
Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1geq 55+3sqrt[3]{27} = 64$$
answered Dec 4 at 21:13
greedoid
36.4k114591
Yup, that first passage was the one I didn't see, I'm actually terrible at AM-GM!...
– Spasoje Durovic
Dec 4 at 21:22
Yeah, I've started working on inequalities about a week ago, hope I'll get better soon! cheers (nice to see you're from the Balkans too :P)
– Spasoje Durovic
Dec 4 at 21:27
Actualy I'm not.
– greedoid
Dec 4 at 21:28
My bad! I supposed it since you used Cyrillic
– Spasoje Durovic
Dec 4 at 21:29
1
Never mind, it is OK
– greedoid
Dec 4 at 21:30
add a comment |
up vote
4
down vote
Since $ab+bc+ca= abc$ and $$ab+bc+cageq 3sqrt[3]{a^2b^2c^2}implies abcgeq 27$$
Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1geq 55+3sqrt[3]{27} = 64$$
answered Dec 4 at 21:13
greedoid
36.4k114591
Yup, that first passage was the one I didn't see, I'm actually terrible at AM-GM!...
– Spasoje Durovic
Dec 4 at 21:22
Yeah, I've started working on inequalities about a week ago, hope I'll get better soon! cheers (nice to see you're from the Balkans too :P)
– Spasoje Durovic
Dec 4 at 21:27
Actualy I'm not.
– greedoid
Dec 4 at 21:28
My bad! I supposed it since you used Cyrillic
– Spasoje Durovic
Dec 4 at 21:29
1
Never mind, it is OK
– greedoid
Dec 4 at 21:30
add a comment |
up vote
4
down vote
up vote
4
down vote
Since $ab+bc+ca= abc$ and $$ab+bc+cageq 3sqrt[3]{a^2b^2c^2}implies abcgeq 27$$
Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1geq 55+3sqrt[3]{27} = 64$$
answered Dec 4 at 21:13
greedoid
36.4k114591
Since $ab+bc+ca= abc$ and $$ab+bc+cageq 3sqrt[3]{a^2b^2c^2}implies abcgeq 27$$
Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1geq 55+3sqrt[3]{27} = 64$$
answered Dec 4 at 21:13
greedoid
36.4k114591
answered Dec 4 at 21:13
greedoid
36.4k114591
answered Dec 4 at 21:13
greedoid
36.4k114591
answered Dec 4 at 21:13
greedoid
36.4k114591
36.4k114591
Yup, that first passage was the one I didn't see, I'm actually terrible at AM-GM!...
– Spasoje Durovic
Dec 4 at 21:22
Yeah, I've started working on inequalities about a week ago, hope I'll get better soon! cheers (nice to see you're from the Balkans too :P)
– Spasoje Durovic
Dec 4 at 21:27
Actualy I'm not.
– greedoid
Dec 4 at 21:28
My bad! I supposed it since you used Cyrillic
– Spasoje Durovic
Dec 4 at 21:29
1
Never mind, it is OK
– greedoid
Dec 4 at 21:30
add a comment |
Yup, that first passage was the one I didn't see, I'm actually terrible at AM-GM!...
– Spasoje Durovic
Dec 4 at 21:22
Yeah, I've started working on inequalities about a week ago, hope I'll get better soon! cheers (nice to see you're from the Balkans too :P)
– Spasoje Durovic
Dec 4 at 21:27
Actualy I'm not.
– greedoid
Dec 4 at 21:28
My bad! I supposed it since you used Cyrillic
– Spasoje Durovic
Dec 4 at 21:29
1
Never mind, it is OK
– greedoid
Dec 4 at 21:30
Yup, that first passage was the one I didn't see, I'm actually terrible at AM-GM!...
– Spasoje Durovic
Dec 4 at 21:22
Yup, that first passage was the one I didn't see, I'm actually terrible at AM-GM!...
– Spasoje Durovic
Dec 4 at 21:22
Yeah, I've started working on inequalities about a week ago, hope I'll get better soon! cheers (nice to see you're from the Balkans too :P)
– Spasoje Durovic
Dec 4 at 21:27
Yeah, I've started working on inequalities about a week ago, hope I'll get better soon! cheers (nice to see you're from the Balkans too :P)
– Spasoje Durovic
Dec 4 at 21:27
Actualy I'm not.
– greedoid
Dec 4 at 21:28
Actualy I'm not.
– greedoid
Dec 4 at 21:28
My bad! I supposed it since you used Cyrillic
– Spasoje Durovic
Dec 4 at 21:29
My bad! I supposed it since you used Cyrillic
– Spasoje Durovic
Dec 4 at 21:29
1
1
Never mind, it is OK
– greedoid
Dec 4 at 21:30
Never mind, it is OK
– greedoid
Dec 4 at 21:30
add a comment |
up vote
2
down vote
For positive variables we need to prove that
$$lnleft((1+a)(1+b)(1+c)right)geqln64$$ or
$$sum_{cyc}(ln(1+a)-2ln2)geq0$$ or
$$sum_{cyc}left(ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)right)geq0,$$
which is true because
$$f(a)=ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)geq0$$ for all $a>0$.
Indeed, $$f'(a)=frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
Can you elaborate on how you got the first expression?
– YiFan
Dec 4 at 22:53
1
@YiFan presumably by taking the $ln(bullet)$ of both sides using the fact that $ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $ln(64)$ term into three equal parts to get it into the cyclic sum, those being $ln(64)/3 = ln(4) = 2 ln(2).$
– CR Drost
Dec 4 at 23:55
@CRDrost Of course. Thanks for the help!
– YiFan
Dec 4 at 23:56
add a comment |
up vote
2
down vote
For positive variables we need to prove that
$$lnleft((1+a)(1+b)(1+c)right)geqln64$$ or
$$sum_{cyc}(ln(1+a)-2ln2)geq0$$ or
$$sum_{cyc}left(ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)right)geq0,$$
which is true because
$$f(a)=ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)geq0$$ for all $a>0$.
Indeed, $$f'(a)=frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
Can you elaborate on how you got the first expression?
– YiFan
Dec 4 at 22:53
1
@YiFan presumably by taking the $ln(bullet)$ of both sides using the fact that $ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $ln(64)$ term into three equal parts to get it into the cyclic sum, those being $ln(64)/3 = ln(4) = 2 ln(2).$
– CR Drost
Dec 4 at 23:55
@CRDrost Of course. Thanks for the help!
– YiFan
Dec 4 at 23:56
add a comment |
up vote
2
down vote
up vote
2
down vote
For positive variables we need to prove that
$$lnleft((1+a)(1+b)(1+c)right)geqln64$$ or
$$sum_{cyc}(ln(1+a)-2ln2)geq0$$ or
$$sum_{cyc}left(ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)right)geq0,$$
which is true because
$$f(a)=ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)geq0$$ for all $a>0$.
Indeed, $$f'(a)=frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
For positive variables we need to prove that
$$lnleft((1+a)(1+b)(1+c)right)geqln64$$ or
$$sum_{cyc}(ln(1+a)-2ln2)geq0$$ or
$$sum_{cyc}left(ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)right)geq0,$$
which is true because
$$f(a)=ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)geq0$$ for all $a>0$.
Indeed, $$f'(a)=frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
edited Dec 5 at 5:26
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
answered Dec 4 at 21:46
Michael Rozenberg
94.8k1588183
94.8k1588183
Can you elaborate on how you got the first expression?
– YiFan
Dec 4 at 22:53
1
@YiFan presumably by taking the $ln(bullet)$ of both sides using the fact that $ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $ln(64)$ term into three equal parts to get it into the cyclic sum, those being $ln(64)/3 = ln(4) = 2 ln(2).$
– CR Drost
Dec 4 at 23:55
@CRDrost Of course. Thanks for the help!
– YiFan
Dec 4 at 23:56
add a comment |
Can you elaborate on how you got the first expression?
– YiFan
Dec 4 at 22:53
1
@YiFan presumably by taking the $ln(bullet)$ of both sides using the fact that $ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $ln(64)$ term into three equal parts to get it into the cyclic sum, those being $ln(64)/3 = ln(4) = 2 ln(2).$
– CR Drost
Dec 4 at 23:55
@CRDrost Of course. Thanks for the help!
– YiFan
Dec 4 at 23:56
Can you elaborate on how you got the first expression?
– YiFan
Dec 4 at 22:53
Can you elaborate on how you got the first expression?
– YiFan
Dec 4 at 22:53
1
1
@YiFan presumably by taking the $ln(bullet)$ of both sides using the fact that $ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $ln(64)$ term into three equal parts to get it into the cyclic sum, those being $ln(64)/3 = ln(4) = 2 ln(2).$
– CR Drost
Dec 4 at 23:55
@YiFan presumably by taking the $ln(bullet)$ of both sides using the fact that $ln$ increases monotonically hence preserving the inequality. Then there is a splitting the of the $ln(64)$ term into three equal parts to get it into the cyclic sum, those being $ln(64)/3 = ln(4) = 2 ln(2).$
– CR Drost
Dec 4 at 23:55
@CRDrost Of course. Thanks for the help!
– YiFan
Dec 4 at 23:56
@CRDrost Of course. Thanks for the help!
– YiFan
Dec 4 at 23:56
add a comment |
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I haven't tried this I admit .... but my instant thought is, since we 'know' (please note the apostrophæ!) that it attains its minimum value at a=b=c=3, could we not hypothesise that it's 3, plug in 3+δ, 3+ε, & 3+ζ, & show that any increase in some measure of the size of δ, ε & ζ causes an increase in the product under the constraint? As I said - just my instantaneous thought on seeing the question!
– AmbretteOrrisey
Dec 4 at 22:54
@AmbretteOrrisey One could take the second partial derivative (since the expression is symetric you can take it with respect to any variable) and see that the function $f(a,b,c)=(a+1)(b+1)(c+1)$ is convex since it's second derivative is equal to zero, given this we basically done what you suggested to (i.e. any increment in a,b,c gives us an increment in f(a,b,c)) but the thing is, you'd still have to formally prove that a=b=c=3 it's the minimal value of the funcion which really doesn't seem easy to me (in one of the answers they actually proved that 3 is the minimal value, you can look it up)
– Spasoje Durovic
Dec 5 at 20:05
@ Spasoje Durovic -- So you can without too much difficulty prove the way said maybe you could? ... assuming that the minimum is at a=b=c=3 ... but then the big stumbilngblock is proving that the minimum is at that? It's funny how so often the stumbilngblock of a proof is the seemingly most trivial part of it!
– AmbretteOrrisey
Dec 5 at 22:03
But do you really need to prove that, logically to prove this theorem as it stands? ... as you can just calculate that at a=b=c=3 the product in the theorem is 64 -- so if you prove that any departure (satisfying the constraint) from that point can only increase that product, surely then you've proven the theorem as it stands? Although I see that if you only have convexity, then you would also need to prove that the minimum is at that particular point.
– AmbretteOrrisey
Dec 5 at 22:17
1
And it looks promising on the face of it! Maybe what you've said then explains why none of the people here have done it that way!
– AmbretteOrrisey
Dec 7 at 15:40