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How can I prove that $(a_1+a_2+dotsb+a_n)(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_n})geq n^2$ [duplicate]

Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$Request for historical precedent of identity implying AM-HM inequalityProve that $frac{a_1^2}{a_1+a_2}+frac{a_2^2}{a_2+a_3}+ cdots frac{a_n^2}{a_n+a_1} geq frac12$How to prove $a_1^m + a_2^m + cdots + a_n^m geq frac{1}{a_1} + frac{1}{a_2} + cdots + frac{1}{a_n}$Prove that: $sqrt[3]{a_1^3+ a_2^3 +cdots+a_n^3} le sqrt{a_1^2 + a_2^2 +cdots+a_n^2}$If $a_igeq 0,$ prove $sumlimits_{n=1}^inftyfrac{a_1+a_2+cdots+a_n}{n}$diverges.$a_1,a_2,…,a_n$ are positive real numbers, their product is equal to $1$, show: $sum_{i=1}^n a_i^{frac 1 i} geq frac{n+1}2$Prove inequality $frac{a_1a_2…a_n}{(a_1+a_2+…+a_n)^n}le frac{(1-a_1)(1-a_2)…(1-a_n)}{(n-a_1-a_2-…-a_n)^n}$Proving the inequality $frac{1}{1+a_1} + frac{2}{(1+a_1)(1+a_2)} + cdots + frac{n}{(1+a_1)ldots (1+a_n)} < 2$Prove by induction $frac{a_1+a_2+cdots+a_n}{n}geq 1$Inequality : $ (a_1a_2+a_2a_3+ldots+a_na_1)left(frac{a_1}{a^2_2+a_2}+frac{a_2}{a^2_3+a_3}+ ldots+frac{a_n}{a^2_1+a_1}right)geq frac{n}{n+1} $How to prove $(a_1+a_2+dots a_n)left(frac{1}{a_1}+frac{1}{a_2}+dots+frac{1}{a_n}right)ge n^2$?

3

1

$begingroup$

This question already has an answer here:

• 
  Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$
  
  7 answers
  
  

I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_n}right)geq n^2$.

Where each $a_i$'s are positive and $n$ is a natural number.

First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if $n=k+1$.

Suppose the inequality holds true when $n=k$, i.e.,

$(a_1+a_2+dotsb+a_k)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}right)geq n^2$.

This is true if and only if

$(a_1+a_2+dotsb+a_k+a_{k+1})left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}+frac{1}{a_{k+1}}right) -a_{k+1}left(frac{1}{a_1}+dotsb+frac{1}{a_k}right)-frac{1}{a_{k+1}}(a_1+dotsb+a_k)-frac{a_{k+1}}{a_{k+1}} geq n^2$.

And I got stuck here.

The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.

analysis inequality

share|cite|improve this question

edited Mar 12 at 20:15

Asaf Karagila♦

306k33438769

asked Mar 12 at 15:02

Ko ByeongminKo Byeongmin

1546

$endgroup$

marked as duplicate by Arnaud D., qwr, Martin R, Asaf Karagila♦ Mar 12 at 20:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Conditions on $a_i$?
  $endgroup$
  – Parcly Taxel
  Mar 12 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  whoa, I forgot the most important info there. They are all positive, and n is a natural number.
  $endgroup$
  – Ko Byeongmin
  Mar 12 at 15:03
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Try Cauchy-Schwarz inequality?
  $endgroup$
  – Sik Feng Cheong
  Mar 12 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now I get it, I learn something new every day!! Thanks a lot :D
  $endgroup$
  – Ko Byeongmin
  Mar 12 at 15:05
  

  
  
  
  

add a comment | 

3

1

$begingroup$

This question already has an answer here:

• 
  Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$
  
  7 answers
  
  

I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_n}right)geq n^2$.

Where each $a_i$'s are positive and $n$ is a natural number.

First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if $n=k+1$.

Suppose the inequality holds true when $n=k$, i.e.,

$(a_1+a_2+dotsb+a_k)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}right)geq n^2$.

This is true if and only if

$(a_1+a_2+dotsb+a_k+a_{k+1})left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}+frac{1}{a_{k+1}}right) -a_{k+1}left(frac{1}{a_1}+dotsb+frac{1}{a_k}right)-frac{1}{a_{k+1}}(a_1+dotsb+a_k)-frac{a_{k+1}}{a_{k+1}} geq n^2$.

And I got stuck here.

The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.

analysis inequality

share|cite|improve this question

edited Mar 12 at 20:15

Asaf Karagila♦

306k33438769

asked Mar 12 at 15:02

Ko ByeongminKo Byeongmin

1546

$endgroup$

marked as duplicate by Arnaud D., qwr, Martin R, Asaf Karagila♦ Mar 12 at 20:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Conditions on $a_i$?
  $endgroup$
  – Parcly Taxel
  Mar 12 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  whoa, I forgot the most important info there. They are all positive, and n is a natural number.
  $endgroup$
  – Ko Byeongmin
  Mar 12 at 15:03
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Try Cauchy-Schwarz inequality?
  $endgroup$
  – Sik Feng Cheong
  Mar 12 at 15:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now I get it, I learn something new every day!! Thanks a lot :D
  $endgroup$
  – Ko Byeongmin
  Mar 12 at 15:05
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$
  
  7 answers
  
  

I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_n}right)geq n^2$.

Where each $a_i$'s are positive and $n$ is a natural number.

First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if $n=k+1$.

Suppose the inequality holds true when $n=k$, i.e.,

$(a_1+a_2+dotsb+a_k)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}right)geq n^2$.

This is true if and only if

$(a_1+a_2+dotsb+a_k+a_{k+1})left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}+frac{1}{a_{k+1}}right) -a_{k+1}left(frac{1}{a_1}+dotsb+frac{1}{a_k}right)-frac{1}{a_{k+1}}(a_1+dotsb+a_k)-frac{a_{k+1}}{a_{k+1}} geq n^2$.

And I got stuck here.

The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.

analysis inequality

share|cite|improve this question

edited Mar 12 at 20:15

Asaf Karagila♦

306k33438769

asked Mar 12 at 15:02

Ko ByeongminKo Byeongmin

1546

$endgroup$

share|cite|improve this question

edited Mar 12 at 20:15

Asaf Karagila♦

306k33438769

asked Mar 12 at 15:02

Ko ByeongminKo Byeongmin

1546

share|cite|improve this question

edited Mar 12 at 20:15

Asaf Karagila♦

306k33438769

asked Mar 12 at 15:02

Ko ByeongminKo Byeongmin

1546

edited Mar 12 at 20:15

Asaf Karagila♦

306k33438769

edited Mar 12 at 20:15

Asaf Karagila♦

306k33438769

asked Mar 12 at 15:02

Ko ByeongminKo Byeongmin

1546

asked Mar 12 at 15:02

Ko ByeongminKo Byeongmin

1546

3 Answers
3

active

oldest

votes

12

$begingroup$

Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]{a_1a_2cdots a_n}
$$ and $$
frac1{a_1}+frac1{a_2}+cdots +frac1{a_n}ge frac{n}{sqrt[n]{a_1a_2cdots a_n}}.
$$

share|cite|improve this answer

answered Mar 12 at 15:06

SongSong

18.3k21549

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's a really strong hint, it looks like an answer
  $endgroup$
  – enedil
  Mar 12 at 16:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
  $endgroup$
  – Song
  Mar 12 at 17:09
  

  
  
  
  

add a comment | 

9

$begingroup$

It is AM-HM inequality
$$frac{a_1+a_2+a_3+...+a_n}{n}geq frac{n}{frac{1}{a_1}+frac{1}{a_2}+frac{1}{a_3}+...+frac{1}{a_n}}$$

share|cite|improve this answer

answered Mar 12 at 15:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

add a comment | 

2

$begingroup$

Here is the proof by induction
that you wanted.

I added a more exact
version of
the identity used
in the proof
at the end.

Let
$s_n
=u_nv_n
$
where
$u_n=sum_{k=1}^n a_k,
v_n= sum_{k=1}^n dfrac1{a_k}
$.
Then,
assuming
$s_n ge n^2$,

$begin{array}\
s_{n+1}
&=u_{n+1}v_{n+1}\
&=(u_n+a_{n+1}) (v_n+dfrac1{a_{n+1}})\
&=u_nv_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&ge n^2+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
end{array}
$

So it is sufficient
to show that
$u_ndfrac1{a_{n+1}}+v_na_{n+1}
ge 2n
$.

By simple algebra,
if $a, b ge 0$ then
$a+b
ge 2sqrt{ab}
$.
(Rewrite as
$(sqrt{a}-sqrt{b})^2ge 0$
or,
as an identity,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$.)

Therefore

$begin{array}\
u_ndfrac1{a_{n+1}}+v_na_{n+1}
&ge sqrt{(u_ndfrac1{a_{n+1}})(v_na_{n+1})}\
&= sqrt{u_nv_n}\
&=2sqrt{s_n}\
&ge 2sqrt{n^2}
qquadtext{by the induction hypothesis}\
&=2n\
end{array}
$

and we are done.

I find it interesting that
$s_n ge n^2$
is used twice in the
induction step.

Note that,
if we use the identity above,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$,
we get this:

$begin{array}\
s_{n+1}
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+2sqrt{u_ndfrac1{a_{n+1}}a_{n+1}v_n}+1+(sqrt{u_ndfrac1{a_{n+1}}}-sqrt{a_{n+1}v_n})^2\
&=s_n+2sqrt{s_n}+1+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&=(sqrt{s_n}+1)^2+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&ge(sqrt{s_n}+1)^2\
end{array}
$

with equality
if and only if
$a_{n+1}
=sqrt{dfrac{u_n}{v_n}}
=sqrt{dfrac{sum_{k=1}^n a_k}{sum_{k=1}^n dfrac1{a_k}}}
$.

share|cite|improve this answer

edited Mar 16 at 20:57

answered Mar 12 at 20:01

marty cohenmarty cohen

74.4k549129

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is wonderful! I love it that you’ve done it without using any “machineries” like the AM-GM inequality. I hope one day I become skillful as you are! Have a great day :)
  $endgroup$
  – Ko Byeongmin
  Mar 12 at 23:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. This makes my day.
  $endgroup$
  – marty cohen
  Mar 13 at 1:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't $a+b =2ab+(sqrt{a}-sqrt{b})^2$ be $a+b =2sqrt{ab}+(sqrt{a}-sqrt{b})^2$?
  $endgroup$
  – Martin R
  Mar 16 at 7:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Thank you. Corrected and comment upvoted.
  $endgroup$
  – marty cohen
  Mar 16 at 20:57
  

  
  
  
  

add a comment | 

12

$begingroup$

Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]{a_1a_2cdots a_n}
$$ and $$
frac1{a_1}+frac1{a_2}+cdots +frac1{a_n}ge frac{n}{sqrt[n]{a_1a_2cdots a_n}}.
$$

share|cite|improve this answer

answered Mar 12 at 15:06

SongSong

18.3k21549

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's a really strong hint, it looks like an answer
  $endgroup$
  – enedil
  Mar 12 at 16:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
  $endgroup$
  – Song
  Mar 12 at 17:09
  

  
  
  
  

add a comment | 

12

$begingroup$

Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]{a_1a_2cdots a_n}
$$ and $$
frac1{a_1}+frac1{a_2}+cdots +frac1{a_n}ge frac{n}{sqrt[n]{a_1a_2cdots a_n}}.
$$

share|cite|improve this answer

answered Mar 12 at 15:06

SongSong

18.3k21549

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That's a really strong hint, it looks like an answer
  $endgroup$
  – enedil
  Mar 12 at 16:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
  $endgroup$
  – Song
  Mar 12 at 17:09
  

  
  
  
  

add a comment | 

$begingroup$

Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]{a_1a_2cdots a_n}
$$ and $$
frac1{a_1}+frac1{a_2}+cdots +frac1{a_n}ge frac{n}{sqrt[n]{a_1a_2cdots a_n}}.
$$

share|cite|improve this answer

answered Mar 12 at 15:06

SongSong

18.3k21549

$endgroup$

share|cite|improve this answer

answered Mar 12 at 15:06

SongSong

18.3k21549

answered Mar 12 at 15:06

SongSong

18.3k21549

answered Mar 12 at 15:06

SongSong

18.3k21549

9

$begingroup$

It is AM-HM inequality
$$frac{a_1+a_2+a_3+...+a_n}{n}geq frac{n}{frac{1}{a_1}+frac{1}{a_2}+frac{1}{a_3}+...+frac{1}{a_n}}$$

share|cite|improve this answer

answered Mar 12 at 15:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

add a comment | 

9

$begingroup$

It is AM-HM inequality
$$frac{a_1+a_2+a_3+...+a_n}{n}geq frac{n}{frac{1}{a_1}+frac{1}{a_2}+frac{1}{a_3}+...+frac{1}{a_n}}$$

share|cite|improve this answer

answered Mar 12 at 15:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

add a comment | 

$begingroup$

It is AM-HM inequality
$$frac{a_1+a_2+a_3+...+a_n}{n}geq frac{n}{frac{1}{a_1}+frac{1}{a_2}+frac{1}{a_3}+...+frac{1}{a_n}}$$

share|cite|improve this answer

answered Mar 12 at 15:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

share|cite|improve this answer

answered Mar 12 at 15:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

answered Mar 12 at 15:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

answered Mar 12 at 15:07

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

2

$begingroup$

Here is the proof by induction
that you wanted.

I added a more exact
version of
the identity used
in the proof
at the end.

Let
$s_n
=u_nv_n
$
where
$u_n=sum_{k=1}^n a_k,
v_n= sum_{k=1}^n dfrac1{a_k}
$.
Then,
assuming
$s_n ge n^2$,

$begin{array}\
s_{n+1}
&=u_{n+1}v_{n+1}\
&=(u_n+a_{n+1}) (v_n+dfrac1{a_{n+1}})\
&=u_nv_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&ge n^2+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
end{array}
$

So it is sufficient
to show that
$u_ndfrac1{a_{n+1}}+v_na_{n+1}
ge 2n
$.

By simple algebra,
if $a, b ge 0$ then
$a+b
ge 2sqrt{ab}
$.
(Rewrite as
$(sqrt{a}-sqrt{b})^2ge 0$
or,
as an identity,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$.)

Therefore

$begin{array}\
u_ndfrac1{a_{n+1}}+v_na_{n+1}
&ge sqrt{(u_ndfrac1{a_{n+1}})(v_na_{n+1})}\
&= sqrt{u_nv_n}\
&=2sqrt{s_n}\
&ge 2sqrt{n^2}
qquadtext{by the induction hypothesis}\
&=2n\
end{array}
$

and we are done.

I find it interesting that
$s_n ge n^2$
is used twice in the
induction step.

Note that,
if we use the identity above,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$,
we get this:

$begin{array}\
s_{n+1}
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+2sqrt{u_ndfrac1{a_{n+1}}a_{n+1}v_n}+1+(sqrt{u_ndfrac1{a_{n+1}}}-sqrt{a_{n+1}v_n})^2\
&=s_n+2sqrt{s_n}+1+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&=(sqrt{s_n}+1)^2+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&ge(sqrt{s_n}+1)^2\
end{array}
$

with equality
if and only if
$a_{n+1}
=sqrt{dfrac{u_n}{v_n}}
=sqrt{dfrac{sum_{k=1}^n a_k}{sum_{k=1}^n dfrac1{a_k}}}
$.

share|cite|improve this answer

edited Mar 16 at 20:57

answered Mar 12 at 20:01

marty cohenmarty cohen

74.4k549129

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is wonderful! I love it that you’ve done it without using any “machineries” like the AM-GM inequality. I hope one day I become skillful as you are! Have a great day :)
  $endgroup$
  – Ko Byeongmin
  Mar 12 at 23:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. This makes my day.
  $endgroup$
  – marty cohen
  Mar 13 at 1:10
  

  
  
  
  


• 
  
  
  

  
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  $begingroup$
  Shouldn't $a+b =2ab+(sqrt{a}-sqrt{b})^2$ be $a+b =2sqrt{ab}+(sqrt{a}-sqrt{b})^2$?
  $endgroup$
  – Martin R
  Mar 16 at 7:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Thank you. Corrected and comment upvoted.
  $endgroup$
  – marty cohen
  Mar 16 at 20:57
  

  
  
  
  

add a comment | 

2

$begingroup$

Here is the proof by induction
that you wanted.

I added a more exact
version of
the identity used
in the proof
at the end.

Let
$s_n
=u_nv_n
$
where
$u_n=sum_{k=1}^n a_k,
v_n= sum_{k=1}^n dfrac1{a_k}
$.
Then,
assuming
$s_n ge n^2$,

$begin{array}\
s_{n+1}
&=u_{n+1}v_{n+1}\
&=(u_n+a_{n+1}) (v_n+dfrac1{a_{n+1}})\
&=u_nv_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&ge n^2+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
end{array}
$

So it is sufficient
to show that
$u_ndfrac1{a_{n+1}}+v_na_{n+1}
ge 2n
$.

By simple algebra,
if $a, b ge 0$ then
$a+b
ge 2sqrt{ab}
$.
(Rewrite as
$(sqrt{a}-sqrt{b})^2ge 0$
or,
as an identity,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$.)

Therefore

$begin{array}\
u_ndfrac1{a_{n+1}}+v_na_{n+1}
&ge sqrt{(u_ndfrac1{a_{n+1}})(v_na_{n+1})}\
&= sqrt{u_nv_n}\
&=2sqrt{s_n}\
&ge 2sqrt{n^2}
qquadtext{by the induction hypothesis}\
&=2n\
end{array}
$

and we are done.

I find it interesting that
$s_n ge n^2$
is used twice in the
induction step.

Note that,
if we use the identity above,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$,
we get this:

$begin{array}\
s_{n+1}
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+2sqrt{u_ndfrac1{a_{n+1}}a_{n+1}v_n}+1+(sqrt{u_ndfrac1{a_{n+1}}}-sqrt{a_{n+1}v_n})^2\
&=s_n+2sqrt{s_n}+1+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&=(sqrt{s_n}+1)^2+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&ge(sqrt{s_n}+1)^2\
end{array}
$

with equality
if and only if
$a_{n+1}
=sqrt{dfrac{u_n}{v_n}}
=sqrt{dfrac{sum_{k=1}^n a_k}{sum_{k=1}^n dfrac1{a_k}}}
$.

share|cite|improve this answer

edited Mar 16 at 20:57

answered Mar 12 at 20:01

marty cohenmarty cohen

74.4k549129

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is wonderful! I love it that you’ve done it without using any “machineries” like the AM-GM inequality. I hope one day I become skillful as you are! Have a great day :)
  $endgroup$
  – Ko Byeongmin
  Mar 12 at 23:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. This makes my day.
  $endgroup$
  – marty cohen
  Mar 13 at 1:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't $a+b =2ab+(sqrt{a}-sqrt{b})^2$ be $a+b =2sqrt{ab}+(sqrt{a}-sqrt{b})^2$?
  $endgroup$
  – Martin R
  Mar 16 at 7:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Thank you. Corrected and comment upvoted.
  $endgroup$
  – marty cohen
  Mar 16 at 20:57
  

  
  
  
  

add a comment | 

$begingroup$

Here is the proof by induction
that you wanted.

I added a more exact
version of
the identity used
in the proof
at the end.

Let
$s_n
=u_nv_n
$
where
$u_n=sum_{k=1}^n a_k,
v_n= sum_{k=1}^n dfrac1{a_k}
$.
Then,
assuming
$s_n ge n^2$,

$begin{array}\
s_{n+1}
&=u_{n+1}v_{n+1}\
&=(u_n+a_{n+1}) (v_n+dfrac1{a_{n+1}})\
&=u_nv_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&ge n^2+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
end{array}
$

So it is sufficient
to show that
$u_ndfrac1{a_{n+1}}+v_na_{n+1}
ge 2n
$.

By simple algebra,
if $a, b ge 0$ then
$a+b
ge 2sqrt{ab}
$.
(Rewrite as
$(sqrt{a}-sqrt{b})^2ge 0$
or,
as an identity,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$.)

Therefore

$begin{array}\
u_ndfrac1{a_{n+1}}+v_na_{n+1}
&ge sqrt{(u_ndfrac1{a_{n+1}})(v_na_{n+1})}\
&= sqrt{u_nv_n}\
&=2sqrt{s_n}\
&ge 2sqrt{n^2}
qquadtext{by the induction hypothesis}\
&=2n\
end{array}
$

and we are done.

I find it interesting that
$s_n ge n^2$
is used twice in the
induction step.

Note that,
if we use the identity above,
$a+b
=2sqrt{ab}+(sqrt{a}-sqrt{b})^2$,
we get this:

$begin{array}\
s_{n+1}
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+2sqrt{u_ndfrac1{a_{n+1}}a_{n+1}v_n}+1+(sqrt{u_ndfrac1{a_{n+1}}}-sqrt{a_{n+1}v_n})^2\
&=s_n+2sqrt{s_n}+1+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&=(sqrt{s_n}+1)^2+dfrac1{a_{n+1}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&ge(sqrt{s_n}+1)^2\
end{array}
$

with equality
if and only if
$a_{n+1}
=sqrt{dfrac{u_n}{v_n}}
=sqrt{dfrac{sum_{k=1}^n a_k}{sum_{k=1}^n dfrac1{a_k}}}
$.

share|cite|improve this answer

edited Mar 16 at 20:57

answered Mar 12 at 20:01

marty cohenmarty cohen

74.4k549129

$endgroup$

share|cite|improve this answer

edited Mar 16 at 20:57

answered Mar 12 at 20:01

marty cohenmarty cohen

74.4k549129

answered Mar 12 at 20:01

marty cohenmarty cohen

74.4k549129

answered Mar 12 at 20:01

marty cohenmarty cohen

74.4k549129