(Proof right now?) Induction Inequality with sum












1














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







share|cite|improve this question
























  • 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
















1














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







share|cite|improve this question
























  • 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














1












1








1







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







share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '18 at 23:16







John M

















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


















  • 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










2 Answers
2






active

oldest

votes


















0














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.






share|cite|improve this answer























  • 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



















0














$a_n$ 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.






share|cite|improve this answer





















  • 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











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2 Answers
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2 Answers
2






active

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votes









0














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.






share|cite|improve this answer























  • 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
















0














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.






share|cite|improve this answer























  • 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














0












0








0






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 22 '18 at 22:51

























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


















  • 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











0














$a_n$ 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.






share|cite|improve this answer





















  • 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
















0














$a_n$ 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.






share|cite|improve this answer





















  • 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














0












0








0






$a_n$ 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.






share|cite|improve this answer












$a_n$ 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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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How to change which sound is reproduced for terminal bell?

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3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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