Proof by induction for tricky double summation: $(sum_{k=1}^n x_k)cdot(sum_{k=1}^n frac{1}{x_k})ge{n^2}$...

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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
  
  

let $x_1....x_n$ be positive integers. Prove by induction the following for natural numbers n:

$(sum_{k=1}^n x_k)cdot(sum_{k=1}^n frac{1}{x_k})ge{n^2}$

Hint: for all positive integers a,b: $frac{a}{b}+frac{b}{a}ge2$

inequality summation induction

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edited Jan 21 at 13:59

Martin Sleziak

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asked Oct 31 '15 at 14:41

Cam MackCam Mack

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marked as duplicate by Martin Sleziak, Martin R, Lord_Farin, Paul Frost, saz Jan 21 at 19:04

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.

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  I highly suspect that OP is supposed to prove that given expression is $geq n^2$.
  $endgroup$
  – Wojowu
  Oct 31 '15 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I'm sorry I missed a part out! I shall fix now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackD'Aurizio I have fixed it now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Wojowu yes you are right
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:38
  

  
  
  
  

add a comment | 

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
  
  

let $x_1....x_n$ be positive integers. Prove by induction the following for natural numbers n:

$(sum_{k=1}^n x_k)cdot(sum_{k=1}^n frac{1}{x_k})ge{n^2}$

Hint: for all positive integers a,b: $frac{a}{b}+frac{b}{a}ge2$

inequality summation induction

share|cite|improve this question

edited Jan 21 at 13:59

Martin Sleziak

44.8k10119273

asked Oct 31 '15 at 14:41

Cam MackCam Mack

185

$endgroup$

marked as duplicate by Martin Sleziak, Martin R, Lord_Farin, Paul Frost, saz Jan 21 at 19:04

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.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I highly suspect that OP is supposed to prove that given expression is $geq n^2$.
  $endgroup$
  – Wojowu
  Oct 31 '15 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I'm sorry I missed a part out! I shall fix now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackD'Aurizio I have fixed it now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Wojowu yes you are right
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:38
  

  
  
  
  

add a comment | 

1

1

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
  
  

let $x_1....x_n$ be positive integers. Prove by induction the following for natural numbers n:

$(sum_{k=1}^n x_k)cdot(sum_{k=1}^n frac{1}{x_k})ge{n^2}$

Hint: for all positive integers a,b: $frac{a}{b}+frac{b}{a}ge2$

inequality summation induction

share|cite|improve this question

edited Jan 21 at 13:59

Martin Sleziak

44.8k10119273

asked Oct 31 '15 at 14:41

Cam MackCam Mack

185

$endgroup$

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
  
  

let $x_1....x_n$ be positive integers. Prove by induction the following for natural numbers n:

$(sum_{k=1}^n x_k)cdot(sum_{k=1}^n frac{1}{x_k})ge{n^2}$

Hint: for all positive integers a,b: $frac{a}{b}+frac{b}{a}ge2$

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
  
  

inequality summation induction

inequality summation induction

share|cite|improve this question

edited Jan 21 at 13:59

Martin Sleziak

44.8k10119273

asked Oct 31 '15 at 14:41

Cam MackCam Mack

185

share|cite|improve this question

edited Jan 21 at 13:59

Martin Sleziak

44.8k10119273

asked Oct 31 '15 at 14:41

Cam MackCam Mack

185

share|cite|improve this question

share|cite|improve this question

edited Jan 21 at 13:59

Martin Sleziak

44.8k10119273

edited Jan 21 at 13:59

Martin Sleziak

44.8k10119273

edited Jan 21 at 13:59

Martin Sleziak

44.8k10119273

44.8k10119273

asked Oct 31 '15 at 14:41

Cam MackCam Mack

185

asked Oct 31 '15 at 14:41

Cam MackCam Mack

185

asked Oct 31 '15 at 14:41

Cam MackCam Mack

185

185

marked as duplicate by Martin Sleziak, Martin R, Lord_Farin, Paul Frost, saz Jan 21 at 19:04

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.

marked as duplicate by Martin Sleziak, Martin R, Lord_Farin, Paul Frost, saz Jan 21 at 19:04

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.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I highly suspect that OP is supposed to prove that given expression is $geq n^2$.
  $endgroup$
  – Wojowu
  Oct 31 '15 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I'm sorry I missed a part out! I shall fix now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackD'Aurizio I have fixed it now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Wojowu yes you are right
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I highly suspect that OP is supposed to prove that given expression is $geq n^2$.
  $endgroup$
  – Wojowu
  Oct 31 '15 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I'm sorry I missed a part out! I shall fix now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JackD'Aurizio I have fixed it now
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Wojowu yes you are right
  $endgroup$
  – Cam Mack
  Oct 31 '15 at 15:38
  

  
  
  
  

1

1

$begingroup$
I highly suspect that OP is supposed to prove that given expression is $geq n^2$.
$endgroup$
– Wojowu
Oct 31 '15 at 14:47

$begingroup$
I highly suspect that OP is supposed to prove that given expression is $geq n^2$.
$endgroup$
– Wojowu
Oct 31 '15 at 14:47

$begingroup$
Oh I'm sorry I missed a part out! I shall fix now
$endgroup$
– Cam Mack
Oct 31 '15 at 15:35

$begingroup$
Oh I'm sorry I missed a part out! I shall fix now
$endgroup$
– Cam Mack
Oct 31 '15 at 15:35

$begingroup$
@JackD'Aurizio I have fixed it now
$endgroup$
– Cam Mack
Oct 31 '15 at 15:37

$begingroup$
@JackD'Aurizio I have fixed it now
$endgroup$
– Cam Mack
Oct 31 '15 at 15:37

$begingroup$
@Wojowu yes you are right
$endgroup$
– Cam Mack
Oct 31 '15 at 15:38

$begingroup$
@Wojowu yes you are right
$endgroup$
– Cam Mack
Oct 31 '15 at 15:38

add a comment | 

1 Answer
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The inductive step should be

$$left(sum_{i=1}^n x_iright)left(sum_{i=1}^n frac{1}{x_i}right)$$
$$=left(sum_{i=1}^{n-1} x_i + x_nright)left(sum_{i=1}^{n-1} frac{1}{x_i} + frac{1}{x_n}right)$$
$$=left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1$$
Now
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) geq (n-1)^2$$
and

$$frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1} frac{1}{x_i}right) = sum_{i=1}^{n-1}left(frac{x_i}{x_n} + frac{x_n}{x_i}right) geq 2(n-1)$$
so that
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1 geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$

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answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will mention that a similar proof can be found in this answer: Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$.
  $endgroup$
  – Martin Sleziak
  Jan 21 at 14:34
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

$begingroup$

The inductive step should be

$$left(sum_{i=1}^n x_iright)left(sum_{i=1}^n frac{1}{x_i}right)$$
$$=left(sum_{i=1}^{n-1} x_i + x_nright)left(sum_{i=1}^{n-1} frac{1}{x_i} + frac{1}{x_n}right)$$
$$=left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1$$
Now
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) geq (n-1)^2$$
and

$$frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1} frac{1}{x_i}right) = sum_{i=1}^{n-1}left(frac{x_i}{x_n} + frac{x_n}{x_i}right) geq 2(n-1)$$
so that
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1 geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$

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answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will mention that a similar proof can be found in this answer: Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$.
  $endgroup$
  – Martin Sleziak
  Jan 21 at 14:34
  

  
  
  
  

add a comment | 

3

$begingroup$

The inductive step should be

$$left(sum_{i=1}^n x_iright)left(sum_{i=1}^n frac{1}{x_i}right)$$
$$=left(sum_{i=1}^{n-1} x_i + x_nright)left(sum_{i=1}^{n-1} frac{1}{x_i} + frac{1}{x_n}right)$$
$$=left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1$$
Now
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) geq (n-1)^2$$
and

$$frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1} frac{1}{x_i}right) = sum_{i=1}^{n-1}left(frac{x_i}{x_n} + frac{x_n}{x_i}right) geq 2(n-1)$$
so that
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1 geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$

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answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will mention that a similar proof can be found in this answer: Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$.
  $endgroup$
  – Martin Sleziak
  Jan 21 at 14:34
  

  
  
  
  

add a comment | 

3

3

3

$begingroup$

The inductive step should be

$$left(sum_{i=1}^n x_iright)left(sum_{i=1}^n frac{1}{x_i}right)$$
$$=left(sum_{i=1}^{n-1} x_i + x_nright)left(sum_{i=1}^{n-1} frac{1}{x_i} + frac{1}{x_n}right)$$
$$=left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1$$
Now
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) geq (n-1)^2$$
and

$$frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1} frac{1}{x_i}right) = sum_{i=1}^{n-1}left(frac{x_i}{x_n} + frac{x_n}{x_i}right) geq 2(n-1)$$
so that
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1 geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$

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answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

$endgroup$

The inductive step should be

$$left(sum_{i=1}^n x_iright)left(sum_{i=1}^n frac{1}{x_i}right)$$
$$=left(sum_{i=1}^{n-1} x_i + x_nright)left(sum_{i=1}^{n-1} frac{1}{x_i} + frac{1}{x_n}right)$$
$$=left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1$$
Now
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) geq (n-1)^2$$
and

$$frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1} frac{1}{x_i}right) = sum_{i=1}^{n-1}left(frac{x_i}{x_n} + frac{x_n}{x_i}right) geq 2(n-1)$$
so that
$$left(sum_{i=1}^{n-1} x_iright)left(sum_{i=1}^{n-1} frac{1}{x_i}right) + frac{1}{x_n}left(sum_{i=1}^{n-1} x_iright) + x_nleft(sum_{i=1}^{n-1}frac{1}{x_i}right) + 1 geq (n-1)^2 + 2(n-1) + 1 = ((n-1) + 1)^2 = n^2$$

share|cite|improve this answer

answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

share|cite|improve this answer

share|cite|improve this answer

answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

answered Oct 31 '15 at 15:47

Chris KerridgeChris Kerridge

1,01137

1,01137

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will mention that a similar proof can be found in this answer: Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$.
  $endgroup$
  – Martin Sleziak
  Jan 21 at 14:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will mention that a similar proof can be found in this answer: Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$.
  $endgroup$
  – Martin Sleziak
  Jan 21 at 14:34
  

  
  
  
  

$begingroup$
I will mention that a similar proof can be found in this answer: Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$.
$endgroup$
– Martin Sleziak
Jan 21 at 14:34

$begingroup$
I will mention that a similar proof can be found in this answer: Proof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$.
$endgroup$
– Martin Sleziak
Jan 21 at 14:34

add a comment | 

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