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


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  • 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












$begingroup$



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











$endgroup$



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














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











$endgroup$





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






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edited Jan 21 at 13:59









Martin Sleziak

44.8k10119273




44.8k10119273










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














  • 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










1 Answer
1






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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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  • $begingroup$
    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


















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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$endgroup$













  • $begingroup$
    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
















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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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






share|cite|improve this answer









$endgroup$



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







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share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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



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