(Proof right now?) Induction Inequality with sum











up vote
1
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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 at 22:14










  • I presume you want each $a_i>0$.
    – Lord Shark the Unknown
    Nov 22 at 22:14










  • $sum a_k$ can be any positive number.
    – Lord Shark the Unknown
    Nov 22 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 at 22:18















up vote
1
down vote

favorite












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 at 22:14










  • I presume you want each $a_i>0$.
    – Lord Shark the Unknown
    Nov 22 at 22:14










  • $sum a_k$ can be any positive number.
    – Lord Shark the Unknown
    Nov 22 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 at 22:18













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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 at 23:16

























asked Nov 22 at 22:11









John M

154




154












  • Shouldn't $a_n$ be positive?
    – Mostafa Ayaz
    Nov 22 at 22:14










  • I presume you want each $a_i>0$.
    – Lord Shark the Unknown
    Nov 22 at 22:14










  • $sum a_k$ can be any positive number.
    – Lord Shark the Unknown
    Nov 22 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 at 22:18


















  • Shouldn't $a_n$ be positive?
    – Mostafa Ayaz
    Nov 22 at 22:14










  • I presume you want each $a_i>0$.
    – Lord Shark the Unknown
    Nov 22 at 22:14










  • $sum a_k$ can be any positive number.
    – Lord Shark the Unknown
    Nov 22 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 at 22:18
















Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 at 22:14




Shouldn't $a_n$ be positive?
– Mostafa Ayaz
Nov 22 at 22:14












I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 at 22:14




I presume you want each $a_i>0$.
– Lord Shark the Unknown
Nov 22 at 22:14












$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 at 22:15




$sum a_k$ can be any positive number.
– Lord Shark the Unknown
Nov 22 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 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 at 22:18










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










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 at 22:45










  • Edited my solution, i think now it should be final, i guess so. Thanks for your help!
    – John M
    Nov 22 at 23:22


















up vote
0
down vote













$a_n$ 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 at 23:23










  • You're welcome. Hope it helps you!
    – Mostafa Ayaz
    Nov 23 at 7:27











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










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 at 22:45










  • Edited my solution, i think now it should be final, i guess so. Thanks for your help!
    – John M
    Nov 22 at 23:22















up vote
0
down vote



accepted










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 at 22:45










  • Edited my solution, i think now it should be final, i guess so. Thanks for your help!
    – John M
    Nov 22 at 23:22













up vote
0
down vote



accepted







up vote
0
down vote



accepted






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 at 22:51

























answered Nov 22 at 22:21









DeepSea

70.8k54487




70.8k54487












  • Thanks for the answer, but i don't know how I should start with this..
    – John M
    Nov 22 at 22:45










  • Edited my solution, i think now it should be final, i guess so. Thanks for your help!
    – John M
    Nov 22 at 23:22


















  • Thanks for the answer, but i don't know how I should start with this..
    – John M
    Nov 22 at 22:45










  • Edited my solution, i think now it should be final, i guess so. Thanks for your help!
    – John M
    Nov 22 at 23:22
















Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 at 22:45




Thanks for the answer, but i don't know how I should start with this..
– John M
Nov 22 at 22:45












Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 at 23:22




Edited my solution, i think now it should be final, i guess so. Thanks for your help!
– John M
Nov 22 at 23:22










up vote
0
down vote













$a_n$ 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 at 23:23










  • You're welcome. Hope it helps you!
    – Mostafa Ayaz
    Nov 23 at 7:27















up vote
0
down vote













$a_n$ 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 at 23:23










  • You're welcome. Hope it helps you!
    – Mostafa Ayaz
    Nov 23 at 7:27













up vote
0
down vote










up vote
0
down vote









$a_n$ 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 at 22:19









Mostafa Ayaz

13.5k3836




13.5k3836












  • Thanks for your help!!
    – John M
    Nov 22 at 23:23










  • You're welcome. Hope it helps you!
    – Mostafa Ayaz
    Nov 23 at 7:27


















  • Thanks for your help!!
    – John M
    Nov 22 at 23:23










  • You're welcome. Hope it helps you!
    – Mostafa Ayaz
    Nov 23 at 7:27
















Thanks for your help!!
– John M
Nov 22 at 23:23




Thanks for your help!!
– John M
Nov 22 at 23:23












You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 at 7:27




You're welcome. Hope it helps you!
– Mostafa Ayaz
Nov 23 at 7:27


















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Bundesstraße 106

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Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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