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Show that:$sum_i=1^n x_i cdot sum_i=1^n frac1x_i geq n^2 $
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Show that:$$sum_i=1^n x_i cdot sum_i=1^n frac1x_i geq n^2 $$
The following hints are also given: $$left(fracxy + fracyx geq 2 right) land x,y gt 0$$
Base Case: For n = 2
$$left(1+2right) cdot left(frac11+frac12right) geq 2^2$$
Inductive hypothesis:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i geq left(n+1right)^2 = n^2+2n+1$$
Inductive step:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i = left(sum_i=1^n x_i+(n+1)right) cdot left(sum_i=1^n frac1x_i+frac1n+1right) = left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) + left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) + (n+1) cdot frac1n+1$$
Final words:
I came to the conclusion that:
$$left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) = n^2$$
I inserted for n = 1 so that $$left((n+1) cdot sum_i=1^n frac1x_iright) = frac21 land left(sum_i=1^n x_i cdot frac1n+1right) = frac12$$ Since $$left(fracxy + fracyx geq 2 right) land x,y$$ was given as a hint in the beginning I thougt I can say that $$left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) = 2n$$
Furthermore it's obvious that: $$(n+1) cdot frac1n+1 = 1$$
It's pretty standard proof by induction and I hope you can maybe give me some advices on what I could have done differently and verify the legitimacy of this proof.
analysis proof-verification induction
asked 3 hours ago
KeJie
1109
add a comment |Â
up vote
4
down vote
favorite
Show that:$$sum_i=1^n x_i cdot sum_i=1^n frac1x_i geq n^2 $$
The following hints are also given: $$left(fracxy + fracyx geq 2 right) land x,y gt 0$$
Base Case: For n = 2
$$left(1+2right) cdot left(frac11+frac12right) geq 2^2$$
Inductive hypothesis:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i geq left(n+1right)^2 = n^2+2n+1$$
Inductive step:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i = left(sum_i=1^n x_i+(n+1)right) cdot left(sum_i=1^n frac1x_i+frac1n+1right) = left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) + left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) + (n+1) cdot frac1n+1$$
Final words:
I came to the conclusion that:
$$left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) = n^2$$
I inserted for n = 1 so that $$left((n+1) cdot sum_i=1^n frac1x_iright) = frac21 land left(sum_i=1^n x_i cdot frac1n+1right) = frac12$$ Since $$left(fracxy + fracyx geq 2 right) land x,y$$ was given as a hint in the beginning I thougt I can say that $$left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) = 2n$$
Furthermore it's obvious that: $$(n+1) cdot frac1n+1 = 1$$
It's pretty standard proof by induction and I hope you can maybe give me some advices on what I could have done differently and verify the legitimacy of this proof.
analysis proof-verification induction
asked 3 hours ago
KeJie
1109
1
Why was this down-voted ?
– Ahmad Bazzi
3 hours ago
1
I don't know but I was a bit shocked when I saw that the question received a downvote 5 seconds after I have just posted it.
– KeJie
3 hours ago
Does $x_i=i$ here?
– Henry
3 hours ago
2
Possible duplicate of Inequality : $sum_k=1^n x_kcdot sum_k=1^n frac1x_k geq n^2$
– StubbornAtom
3 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Show that:$$sum_i=1^n x_i cdot sum_i=1^n frac1x_i geq n^2 $$
The following hints are also given: $$left(fracxy + fracyx geq 2 right) land x,y gt 0$$
Base Case: For n = 2
$$left(1+2right) cdot left(frac11+frac12right) geq 2^2$$
Inductive hypothesis:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i geq left(n+1right)^2 = n^2+2n+1$$
Inductive step:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i = left(sum_i=1^n x_i+(n+1)right) cdot left(sum_i=1^n frac1x_i+frac1n+1right) = left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) + left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) + (n+1) cdot frac1n+1$$
Final words:
I came to the conclusion that:
$$left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) = n^2$$
I inserted for n = 1 so that $$left((n+1) cdot sum_i=1^n frac1x_iright) = frac21 land left(sum_i=1^n x_i cdot frac1n+1right) = frac12$$ Since $$left(fracxy + fracyx geq 2 right) land x,y$$ was given as a hint in the beginning I thougt I can say that $$left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) = 2n$$
Furthermore it's obvious that: $$(n+1) cdot frac1n+1 = 1$$
It's pretty standard proof by induction and I hope you can maybe give me some advices on what I could have done differently and verify the legitimacy of this proof.
analysis proof-verification induction
asked 3 hours ago
KeJie
1109
Show that:$$sum_i=1^n x_i cdot sum_i=1^n frac1x_i geq n^2 $$
The following hints are also given: $$left(fracxy + fracyx geq 2 right) land x,y gt 0$$
Base Case: For n = 2
$$left(1+2right) cdot left(frac11+frac12right) geq 2^2$$
Inductive hypothesis:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i geq left(n+1right)^2 = n^2+2n+1$$
Inductive step:
$$sum_i=1^n+1 x_i cdot sum_i=1^n+1 frac1x_i = left(sum_i=1^n x_i+(n+1)right) cdot left(sum_i=1^n frac1x_i+frac1n+1right) = left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) + left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) + (n+1) cdot frac1n+1$$
Final words:
I came to the conclusion that:
$$left(sum_i=1^n x_i cdot sum_i=1^n frac1x_iright) = n^2$$
I inserted for n = 1 so that $$left((n+1) cdot sum_i=1^n frac1x_iright) = frac21 land left(sum_i=1^n x_i cdot frac1n+1right) = frac12$$ Since $$left(fracxy + fracyx geq 2 right) land x,y$$ was given as a hint in the beginning I thougt I can say that $$left((n+1) cdot sum_i=1^n frac1x_iright) + left(sum_i=1^n x_i cdot frac1n+1right) = 2n$$
Furthermore it's obvious that: $$(n+1) cdot frac1n+1 = 1$$
It's pretty standard proof by induction and I hope you can maybe give me some advices on what I could have done differently and verify the legitimacy of this proof.
analysis proof-verification induction
asked 3 hours ago
KeJie
1109
asked 3 hours ago
KeJie
1109
asked 3 hours ago
KeJie
1109
asked 3 hours ago
KeJie
1109
1109
1
Why was this down-voted ?
– Ahmad Bazzi
3 hours ago
1
I don't know but I was a bit shocked when I saw that the question received a downvote 5 seconds after I have just posted it.
– KeJie
3 hours ago
Does $x_i=i$ here?
– Henry
3 hours ago
2
Possible duplicate of Inequality : $sum_k=1^n x_kcdot sum_k=1^n frac1x_k geq n^2$
– StubbornAtom
3 hours ago
add a comment |Â
1
Why was this down-voted ?
– Ahmad Bazzi
3 hours ago
1
I don't know but I was a bit shocked when I saw that the question received a downvote 5 seconds after I have just posted it.
– KeJie
3 hours ago
Does $x_i=i$ here?
– Henry
3 hours ago
2
Possible duplicate of Inequality : $sum_k=1^n x_kcdot sum_k=1^n frac1x_k geq n^2$
– StubbornAtom
3 hours ago
1
1
Why was this down-voted ?
– Ahmad Bazzi
3 hours ago
Why was this down-voted ?
– Ahmad Bazzi
3 hours ago
1
1
I don't know but I was a bit shocked when I saw that the question received a downvote 5 seconds after I have just posted it.
– KeJie
3 hours ago
I don't know but I was a bit shocked when I saw that the question received a downvote 5 seconds after I have just posted it.
– KeJie
3 hours ago
Does $x_i=i$ here?
– Henry
3 hours ago
Does $x_i=i$ here?
– Henry
3 hours ago
2
2
Possible duplicate of Inequality : $sum_k=1^n x_kcdot sum_k=1^n frac1x_k geq n^2$
– StubbornAtom
3 hours ago
Possible duplicate of Inequality : $sum_k=1^n x_kcdot sum_k=1^n frac1x_k geq n^2$
– StubbornAtom
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Alternative way :
It's a direct consequence of the inequality of Chebyshev .
answered 3 hours ago
max8128
765320
Thank you for sharing it I was not aware of that.
– KeJie
3 hours ago
add a comment |Â
up vote
0
down vote
One must assume $,x_i gt 0,$ otherwise the inequality doesn't necessarily hold true.
Once assumed that all $,x_i,$ are positive, this is just the AM $ge$ HM mean inequality:
$$fracsum_i=1^n x_in ;ge; fracnsum_i=1^n frac1x_i$$
answered 1 hour ago
dxiv
53.6k64696
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Alternative way :
It's a direct consequence of the inequality of Chebyshev .
answered 3 hours ago
max8128
765320
Thank you for sharing it I was not aware of that.
– KeJie
3 hours ago
add a comment |Â
up vote
1
down vote
Alternative way :
It's a direct consequence of the inequality of Chebyshev .
answered 3 hours ago
max8128
765320
Thank you for sharing it I was not aware of that.
– KeJie
3 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Alternative way :
It's a direct consequence of the inequality of Chebyshev .
answered 3 hours ago
max8128
765320
Alternative way :
It's a direct consequence of the inequality of Chebyshev .
answered 3 hours ago
max8128
765320
answered 3 hours ago
max8128
765320
answered 3 hours ago
max8128
765320
answered 3 hours ago
max8128
765320
765320
Thank you for sharing it I was not aware of that.
– KeJie
3 hours ago
add a comment |Â
Thank you for sharing it I was not aware of that.
– KeJie
3 hours ago
Thank you for sharing it I was not aware of that.
– KeJie
3 hours ago
Thank you for sharing it I was not aware of that.
– KeJie
3 hours ago
add a comment |Â
up vote
0
down vote
One must assume $,x_i gt 0,$ otherwise the inequality doesn't necessarily hold true.
Once assumed that all $,x_i,$ are positive, this is just the AM $ge$ HM mean inequality:
$$fracsum_i=1^n x_in ;ge; fracnsum_i=1^n frac1x_i$$
answered 1 hour ago
dxiv
53.6k64696
add a comment |Â
up vote
0
down vote
One must assume $,x_i gt 0,$ otherwise the inequality doesn't necessarily hold true.
Once assumed that all $,x_i,$ are positive, this is just the AM $ge$ HM mean inequality:
$$fracsum_i=1^n x_in ;ge; fracnsum_i=1^n frac1x_i$$
answered 1 hour ago
dxiv
53.6k64696
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One must assume $,x_i gt 0,$ otherwise the inequality doesn't necessarily hold true.
Once assumed that all $,x_i,$ are positive, this is just the AM $ge$ HM mean inequality:
$$fracsum_i=1^n x_in ;ge; fracnsum_i=1^n frac1x_i$$
answered 1 hour ago
dxiv
53.6k64696
One must assume $,x_i gt 0,$ otherwise the inequality doesn't necessarily hold true.
Once assumed that all $,x_i,$ are positive, this is just the AM $ge$ HM mean inequality:
$$fracsum_i=1^n x_in ;ge; fracnsum_i=1^n frac1x_i$$
answered 1 hour ago
dxiv
53.6k64696
answered 1 hour ago
dxiv
53.6k64696
answered 1 hour ago
dxiv
53.6k64696
answered 1 hour ago
dxiv
53.6k64696
53.6k64696
add a comment |Â
add a comment |Â
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1
Why was this down-voted ?
– Ahmad Bazzi
3 hours ago
1
I don't know but I was a bit shocked when I saw that the question received a downvote 5 seconds after I have just posted it.
– KeJie
3 hours ago
Does $x_i=i$ here?
– Henry
3 hours ago
2
Possible duplicate of Inequality : $sum_k=1^n x_kcdot sum_k=1^n frac1x_k geq n^2$
– StubbornAtom
3 hours ago