Tricky AM-GM inequalityProof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$How to use induction on this type of inequality?How to prove $a_1^m + a_2^m + cdots + a_n^m geq frac1a_1 + frac1a_2 + cdots + frac1a_n$Bernoulli's inequality variationInequality $( sum_i=1^n a_i)( sum_i=1^n frac1a_i)geq n^2$Proof of this inequalityRearrangement Inequality Problem: Mathematical Olympiad.General inequalityInequality : $ (a_1a_2+a_2a_3+ldots+a_na_1)left(fraca_1a^2_2+a_2+fraca_2a^2_3+a_3+ ldots+fraca_na^2_1+a_1right)geq fracnn+1 $Regarding AM-GM inequalityInequality with two sums
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Tricky AM-GM inequality
Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$How to use induction on this type of inequality?How to prove $a_1^m + a_2^m + cdots + a_n^m geq frac1a_1 + frac1a_2 + cdots + frac1a_n$Bernoulli's inequality variationInequality $( sum_i=1^n a_i)( sum_i=1^n frac1a_i)geq n^2$Proof of this inequalityRearrangement Inequality Problem: Mathematical Olympiad.General inequalityInequality : $ (a_1a_2+a_2a_3+ldots+a_na_1)left(fraca_1a^2_2+a_2+fraca_2a^2_3+a_3+ ldots+fraca_na^2_1+a_1right)geq fracnn+1 $Regarding AM-GM inequalityInequality with two sums
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I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.
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.,
$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.
This is true if and only if
$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.
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
$endgroup$
add a comment |
$begingroup$
I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.
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.,
$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.
This is true if and only if
$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.
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
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$begingroup$
Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago
$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago
4
$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago
$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago
2
$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago
add a comment |
$begingroup$
I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.
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.,
$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.
This is true if and only if
$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.
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
$endgroup$
I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac1a_1+frac1a_2+dotsb+frac1a_nright)geq n^2$.
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.,
$(a_1+a_2+dotsb+a_k)left(frac1a_1+frac1a_2+dotsb+frac1a_kright)geq n^2$.
This is true if and only if
$(a_1+a_2+dotsb+a_k+a_k+1)left(frac1a_1+frac1a_2+dotsb+frac1a_k+frac1a_k+1right) -a_k+1left(frac1a_1+dotsb+frac1a_kright)-frac1a_k+1(a_1+dotsb+a_k)-fraca_k+1a_k+1 geq n^2$.
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
analysis inequality
edited 4 hours ago
Ko Byeongmin
asked 4 hours ago
Ko ByeongminKo Byeongmin
1546
1546
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Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago
$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago
4
$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago
$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago
2
$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago
add a comment |
$begingroup$
Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago
$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago
4
$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago
$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago
2
$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago
$begingroup$
Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago
$begingroup$
Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago
$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago
$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago
4
4
$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago
$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago
$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago
$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago
2
2
$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago
$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
$$ and $$
frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
$$
$endgroup$
$begingroup$
That's a really strong hint, it looks like an answer
$endgroup$
– enedil
2 hours ago
$begingroup$
You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
$endgroup$
– Song
2 hours ago
add a comment |
$begingroup$
It is AM-HM inequality
$$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
$$ and $$
frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
$$
$endgroup$
$begingroup$
That's a really strong hint, it looks like an answer
$endgroup$
– enedil
2 hours ago
$begingroup$
You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
$endgroup$
– Song
2 hours ago
add a comment |
$begingroup$
Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
$$ and $$
frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
$$
$endgroup$
$begingroup$
That's a really strong hint, it looks like an answer
$endgroup$
– enedil
2 hours ago
$begingroup$
You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
$endgroup$
– Song
2 hours ago
add a comment |
$begingroup$
Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
$$ and $$
frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
$$
$endgroup$
Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]a_1a_2cdots a_n
$$ and $$
frac1a_1+frac1a_2+cdots +frac1a_nge fracnsqrt[n]a_1a_2cdots a_n.
$$
answered 4 hours ago
SongSong
17.2k21246
17.2k21246
$begingroup$
That's a really strong hint, it looks like an answer
$endgroup$
– enedil
2 hours ago
$begingroup$
You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
$endgroup$
– Song
2 hours ago
add a comment |
$begingroup$
That's a really strong hint, it looks like an answer
$endgroup$
– enedil
2 hours ago
$begingroup$
You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
$endgroup$
– Song
2 hours ago
$begingroup$
That's a really strong hint, it looks like an answer
$endgroup$
– enedil
2 hours ago
$begingroup$
That's a really strong hint, it looks like an answer
$endgroup$
– enedil
2 hours ago
$begingroup$
You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
$endgroup$
– Song
2 hours ago
$begingroup$
You may be right.. It makes the remaining step look so trivial, so it can be justly regarded as an answer.
$endgroup$
– Song
2 hours ago
add a comment |
$begingroup$
It is AM-HM inequality
$$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$
$endgroup$
add a comment |
$begingroup$
It is AM-HM inequality
$$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$
$endgroup$
add a comment |
$begingroup$
It is AM-HM inequality
$$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$
$endgroup$
It is AM-HM inequality
$$fraca_1+a_2+a_3+...+a_nngeq fracnfrac1a_1+frac1a_2+frac1a_3+...+frac1a_n$$
answered 4 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
77.7k42866
add a comment |
add a comment |
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-analysis, inequality
$begingroup$
Conditions on $a_i$?
$endgroup$
– Parcly Taxel
4 hours ago
$begingroup$
whoa, I forgot the most important info there. They are all positive, and n is a natural number.
$endgroup$
– Ko Byeongmin
4 hours ago
4
$begingroup$
Try Cauchy-Schwarz inequality?
$endgroup$
– Sik Feng Cheong
4 hours ago
$begingroup$
Now I get it, I learn something new every day!! Thanks a lot :D
$endgroup$
– Ko Byeongmin
4 hours ago
2
$begingroup$
Possible duplicate of Proof that $left(sum^n_k=1x_kright)left(sum^n_k=1y_kright)geq n^2$
$endgroup$
– Arnaud D.
3 hours ago