(Proof right now?) Induction Inequality with sum
Have a short question to this equation.
I have to proof it per induction.
$$sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k} geq n^2 = ncdot n$$
...
so:
For induction proof can I use this? Is the following step correct?
$$sum_{k=1}^n a_k geq n$$
$$sum_{k=1}^n dfrac{1}{a_k} geq n$$
Thanks for your answers!
--EDIT--
I should proof that:
$$sum_{k=1}^{n+1} a_kcdotsum_{k=1}^{n+1} dfrac{1}{a_k} geq (n+1)^2 = n^2+2n+1$$
At the end of my proof there are following lines:
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) + left(dfrac{1}{a_{n+1}}cdot sum_{k=1}^n a_kright) +left((a_{n+1}) cdot sum_{k=1}^n dfrac{1}{a_k}right) + 1 $$
Now we can see that this part
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) +1 geq n^2+1$$
But what about the other 2 parts?
--- NEW STEPS ---
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
so now I substitute these:
$$left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right)geq 2n$$
induction start:
...
Induction end:
$$left(sum_{k=1}^{n+1} b_k+ dfrac{1}{b_k}right) = left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) + left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) $$
$$ left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) geq 2n$$
$$ left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) geq 2$$
So the whole of this:
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
and now the first equation is proofed.
Is this right, and finally the end? :)
analysis induction
asked Nov 22 '18 at 22:11
John MJohn M
154
add a comment |
Have a short question to this equation.
I have to proof it per induction.
$$sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k} geq n^2 = ncdot n$$
...
so:
For induction proof can I use this? Is the following step correct?
$$sum_{k=1}^n a_k geq n$$
$$sum_{k=1}^n dfrac{1}{a_k} geq n$$
Thanks for your answers!
--EDIT--
I should proof that:
$$sum_{k=1}^{n+1} a_kcdotsum_{k=1}^{n+1} dfrac{1}{a_k} geq (n+1)^2 = n^2+2n+1$$
At the end of my proof there are following lines:
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) + left(dfrac{1}{a_{n+1}}cdot sum_{k=1}^n a_kright) +left((a_{n+1}) cdot sum_{k=1}^n dfrac{1}{a_k}right) + 1 $$
Now we can see that this part
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) +1 geq n^2+1$$
But what about the other 2 parts?
--- NEW STEPS ---
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
so now I substitute these:
$$left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right)geq 2n$$
induction start:
...
Induction end:
$$left(sum_{k=1}^{n+1} b_k+ dfrac{1}{b_k}right) = left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) + left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) $$
$$ left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) geq 2n$$
$$ left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) geq 2$$
So the whole of this:
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
and now the first equation is proofed.
Is this right, and finally the end? :)
analysis induction
asked Nov 22 '18 at 22:11
John MJohn M
154
Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 '18 at 22:14
I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 '18 at 22:14
$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 '18 at 22:15
Yeah it can be any positive number. Can i assume in this inequality that the one sum is bigger than n and the other sum is bigger than n?
– John M
Nov 22 '18 at 22:18
add a comment |
Have a short question to this equation.
I have to proof it per induction.
$$sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k} geq n^2 = ncdot n$$
...
so:
For induction proof can I use this? Is the following step correct?
$$sum_{k=1}^n a_k geq n$$
$$sum_{k=1}^n dfrac{1}{a_k} geq n$$
Thanks for your answers!
--EDIT--
I should proof that:
$$sum_{k=1}^{n+1} a_kcdotsum_{k=1}^{n+1} dfrac{1}{a_k} geq (n+1)^2 = n^2+2n+1$$
At the end of my proof there are following lines:
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) + left(dfrac{1}{a_{n+1}}cdot sum_{k=1}^n a_kright) +left((a_{n+1}) cdot sum_{k=1}^n dfrac{1}{a_k}right) + 1 $$
Now we can see that this part
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) +1 geq n^2+1$$
But what about the other 2 parts?
--- NEW STEPS ---
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
so now I substitute these:
$$left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right)geq 2n$$
induction start:
...
Induction end:
$$left(sum_{k=1}^{n+1} b_k+ dfrac{1}{b_k}right) = left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) + left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) $$
$$ left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) geq 2n$$
$$ left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) geq 2$$
So the whole of this:
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
and now the first equation is proofed.
Is this right, and finally the end? :)
analysis induction
asked Nov 22 '18 at 22:11
John MJohn M
154
Have a short question to this equation.
I have to proof it per induction.
$$sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k} geq n^2 = ncdot n$$
...
so:
For induction proof can I use this? Is the following step correct?
$$sum_{k=1}^n a_k geq n$$
$$sum_{k=1}^n dfrac{1}{a_k} geq n$$
Thanks for your answers!
--EDIT--
I should proof that:
$$sum_{k=1}^{n+1} a_kcdotsum_{k=1}^{n+1} dfrac{1}{a_k} geq (n+1)^2 = n^2+2n+1$$
At the end of my proof there are following lines:
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) + left(dfrac{1}{a_{n+1}}cdot sum_{k=1}^n a_kright) +left((a_{n+1}) cdot sum_{k=1}^n dfrac{1}{a_k}right) + 1 $$
Now we can see that this part
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) +1 geq n^2+1$$
But what about the other 2 parts?
--- NEW STEPS ---
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
so now I substitute these:
$$left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right)geq 2n$$
induction start:
...
Induction end:
$$left(sum_{k=1}^{n+1} b_k+ dfrac{1}{b_k}right) = left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) + left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) $$
$$ left(sum_{k=1}^n b_k+ dfrac{1}{b_k}right) geq 2n$$
$$ left(sum_{k=n+1}^{n+1} b_k+ dfrac{1}{b_k}right) geq 2$$
So the whole of this:
$$left(sum_{k=1}^n dfrac{a_k}{a_{n+1}}+ dfrac{a_{n+1}}{a_k}right)geq 2n$$
and now the first equation is proofed.
Is this right, and finally the end? :)
analysis induction
analysis induction
asked Nov 22 '18 at 22:11
John MJohn M
154
asked Nov 22 '18 at 22:11
John MJohn M
154
edited Nov 22 '18 at 23:16
John M
asked Nov 22 '18 at 22:11
John MJohn M
154
asked Nov 22 '18 at 22:11
John MJohn M
154
asked Nov 22 '18 at 22:11
John MJohn M
154
154
Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 '18 at 22:14
I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 '18 at 22:14
$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 '18 at 22:15
Yeah it can be any positive number. Can i assume in this inequality that the one sum is bigger than n and the other sum is bigger than n?
– John M
Nov 22 '18 at 22:18
add a comment |
Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 '18 at 22:14
I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 '18 at 22:14
$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 '18 at 22:15
Yeah it can be any positive number. Can i assume in this inequality that the one sum is bigger than n and the other sum is bigger than n?
– John M
Nov 22 '18 at 22:18
Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 '18 at 22:14
Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 '18 at 22:14
I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 '18 at 22:14
I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 '18 at 22:14
$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 '18 at 22:15
$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 '18 at 22:15
Yeah it can be any positive number. Can i assume in this inequality that the one sum is bigger than n and the other sum is bigger than n?
– John M
Nov 22 '18 at 22:18
Yeah it can be any positive number. Can i assume in this inequality that the one sum is bigger than n and the other sum is bigger than n?
– John M
Nov 22 '18 at 22:18
add a comment |
2 Answers
2
active
oldest
votes
Hint: In the inductive step, observe that $displaystyle sum_{k=1}^nleft(dfrac{a_{n+1}}{a_k}+dfrac{a_k}{a_{n+1}}right)ge 2n$ by AM-GM inequality, and together with the inductive step we're done.I want to point out that in using your answer $LHS =$ inductive part $+$ the two parts $+1ge n^2+2n+1=(n+1)^2=RHS$ ,completing the proof.
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 '18 at 22:45
Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 '18 at 23:22
add a comment |
$a_n$ must be always positive. Here is a better way to attain to this:$$sum_{k=1}^{n}a_ksum_{l=1}^{n}{1over a_l}=sum_{k,l}{a_kover a_l}=sum_{k<l}{a_kover a_l}+sum_{k>l}{a_kover a_l}+sum_{k=l}{a_kover a_l}=n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)$$also we know that $${xover y}+{yover x}ge2$$therefore$$n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)ge n+sum_{k<l}2=n+2cdot {n^2-nover 2}=n^2$$which completes the proof.
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
Thanks for your help!!
– John M
Nov 22 '18 at 23:23
You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 '18 at 7:27
add a comment |
Your Answer
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2 Answers
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2 Answers
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Hint: In the inductive step, observe that $displaystyle sum_{k=1}^nleft(dfrac{a_{n+1}}{a_k}+dfrac{a_k}{a_{n+1}}right)ge 2n$ by AM-GM inequality, and together with the inductive step we're done.I want to point out that in using your answer $LHS =$ inductive part $+$ the two parts $+1ge n^2+2n+1=(n+1)^2=RHS$ ,completing the proof.
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 '18 at 22:45
Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 '18 at 23:22
add a comment |
Hint: In the inductive step, observe that $displaystyle sum_{k=1}^nleft(dfrac{a_{n+1}}{a_k}+dfrac{a_k}{a_{n+1}}right)ge 2n$ by AM-GM inequality, and together with the inductive step we're done.I want to point out that in using your answer $LHS =$ inductive part $+$ the two parts $+1ge n^2+2n+1=(n+1)^2=RHS$ ,completing the proof.
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 '18 at 22:45
Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 '18 at 23:22
add a comment |
Hint: In the inductive step, observe that $displaystyle sum_{k=1}^nleft(dfrac{a_{n+1}}{a_k}+dfrac{a_k}{a_{n+1}}right)ge 2n$ by AM-GM inequality, and together with the inductive step we're done.I want to point out that in using your answer $LHS =$ inductive part $+$ the two parts $+1ge n^2+2n+1=(n+1)^2=RHS$ ,completing the proof.
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
Hint: In the inductive step, observe that $displaystyle sum_{k=1}^nleft(dfrac{a_{n+1}}{a_k}+dfrac{a_k}{a_{n+1}}right)ge 2n$ by AM-GM inequality, and together with the inductive step we're done.I want to point out that in using your answer $LHS =$ inductive part $+$ the two parts $+1ge n^2+2n+1=(n+1)^2=RHS$ ,completing the proof.
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
edited Nov 22 '18 at 22:51
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
answered Nov 22 '18 at 22:21
DeepSeaDeepSea
70.9k54487
70.9k54487
Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 '18 at 22:45
Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 '18 at 23:22
add a comment |
Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 '18 at 22:45
Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 '18 at 23:22
Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 '18 at 22:45
Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 '18 at 22:45
Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 '18 at 23:22
Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 '18 at 23:22
add a comment |
$a_n$ must be always positive. Here is a better way to attain to this:$$sum_{k=1}^{n}a_ksum_{l=1}^{n}{1over a_l}=sum_{k,l}{a_kover a_l}=sum_{k<l}{a_kover a_l}+sum_{k>l}{a_kover a_l}+sum_{k=l}{a_kover a_l}=n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)$$also we know that $${xover y}+{yover x}ge2$$therefore$$n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)ge n+sum_{k<l}2=n+2cdot {n^2-nover 2}=n^2$$which completes the proof.
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
Thanks for your help!!
– John M
Nov 22 '18 at 23:23
You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 '18 at 7:27
add a comment |
$a_n$ must be always positive. Here is a better way to attain to this:$$sum_{k=1}^{n}a_ksum_{l=1}^{n}{1over a_l}=sum_{k,l}{a_kover a_l}=sum_{k<l}{a_kover a_l}+sum_{k>l}{a_kover a_l}+sum_{k=l}{a_kover a_l}=n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)$$also we know that $${xover y}+{yover x}ge2$$therefore$$n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)ge n+sum_{k<l}2=n+2cdot {n^2-nover 2}=n^2$$which completes the proof.
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
Thanks for your help!!
– John M
Nov 22 '18 at 23:23
You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 '18 at 7:27
add a comment |
$a_n$ must be always positive. Here is a better way to attain to this:$$sum_{k=1}^{n}a_ksum_{l=1}^{n}{1over a_l}=sum_{k,l}{a_kover a_l}=sum_{k<l}{a_kover a_l}+sum_{k>l}{a_kover a_l}+sum_{k=l}{a_kover a_l}=n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)$$also we know that $${xover y}+{yover x}ge2$$therefore$$n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)ge n+sum_{k<l}2=n+2cdot {n^2-nover 2}=n^2$$which completes the proof.
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
$a_n$ must be always positive. Here is a better way to attain to this:$$sum_{k=1}^{n}a_ksum_{l=1}^{n}{1over a_l}=sum_{k,l}{a_kover a_l}=sum_{k<l}{a_kover a_l}+sum_{k>l}{a_kover a_l}+sum_{k=l}{a_kover a_l}=n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)$$also we know that $${xover y}+{yover x}ge2$$therefore$$n+sum_{k<l}left({a_kover a_l}+{a_lover a_k} right)ge n+sum_{k<l}2=n+2cdot {n^2-nover 2}=n^2$$which completes the proof.
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
answered Nov 22 '18 at 22:19
Mostafa AyazMostafa Ayaz
14.2k3937
14.2k3937
Thanks for your help!!
– John M
Nov 22 '18 at 23:23
You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 '18 at 7:27
add a comment |
Thanks for your help!!
– John M
Nov 22 '18 at 23:23
You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 '18 at 7:27
Thanks for your help!!
– John M
Nov 22 '18 at 23:23
Thanks for your help!!
– John M
Nov 22 '18 at 23:23
You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 '18 at 7:27
You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 '18 at 7:27
add a comment |
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Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 '18 at 22:14
I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 '18 at 22:14
$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 '18 at 22:15
Yeah it can be any positive number. Can i assume in this inequality that the one sum is bigger than n and the other sum is bigger than n?
– John M
Nov 22 '18 at 22:18