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Prove an inequality using the Cauchy Schwarz inequality
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How do I prove this inequality
$$ 1<frac1n+1+frac1n+2+ ... + frac13n+1<2 $$
I've been thinking of proving that each term is bigger than a term of another sequence and that the sum of that sequence is greater than 1 and the other way around to prove that the sum is less than 2.
inequality cauchy-schwarz-inequality
edited 1 hour ago
Zvi
660111
asked 2 hours ago
Diana
264
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Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
3
down vote
favorite
How do I prove this inequality
$$ 1<frac1n+1+frac1n+2+ ... + frac13n+1<2 $$
I've been thinking of proving that each term is bigger than a term of another sequence and that the sum of that sequence is greater than 1 and the other way around to prove that the sum is less than 2.
inequality cauchy-schwarz-inequality
edited 1 hour ago
Zvi
660111
asked 2 hours ago
Diana
264
New contributor
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Are you sure you need Cauchy-Schwarz inequality? It can be solved by simple algebraic manipulation.
– Gibbs
1 hour ago
1
The teacher asked us to try to solve it with Cauchy-Schwarz
– Diana
1 hour ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How do I prove this inequality
$$ 1<frac1n+1+frac1n+2+ ... + frac13n+1<2 $$
I've been thinking of proving that each term is bigger than a term of another sequence and that the sum of that sequence is greater than 1 and the other way around to prove that the sum is less than 2.
inequality cauchy-schwarz-inequality
edited 1 hour ago
Zvi
660111
asked 2 hours ago
Diana
264
New contributor
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How do I prove this inequality
$$ 1<frac1n+1+frac1n+2+ ... + frac13n+1<2 $$
I've been thinking of proving that each term is bigger than a term of another sequence and that the sum of that sequence is greater than 1 and the other way around to prove that the sum is less than 2.
inequality cauchy-schwarz-inequality
inequality cauchy-schwarz-inequality
edited 1 hour ago
Zvi
660111
asked 2 hours ago
Diana
264
New contributor
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Zvi
660111
asked 2 hours ago
Diana
264
New contributor
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Zvi
660111
edited 1 hour ago
Zvi
660111
edited 1 hour ago
Zvi
660111
660111
asked 2 hours ago
Diana
264
New contributor
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Diana
264
asked 2 hours ago
Diana
264
264
New contributor
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Diana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Are you sure you need Cauchy-Schwarz inequality? It can be solved by simple algebraic manipulation.
– Gibbs
1 hour ago
1
The teacher asked us to try to solve it with Cauchy-Schwarz
– Diana
1 hour ago
add a comment |Â
Are you sure you need Cauchy-Schwarz inequality? It can be solved by simple algebraic manipulation.
– Gibbs
1 hour ago
1
The teacher asked us to try to solve it with Cauchy-Schwarz
– Diana
1 hour ago
Are you sure you need Cauchy-Schwarz inequality? It can be solved by simple algebraic manipulation.
– Gibbs
1 hour ago
Are you sure you need Cauchy-Schwarz inequality? It can be solved by simple algebraic manipulation.
– Gibbs
1 hour ago
1
1
The teacher asked us to try to solve it with Cauchy-Schwarz
– Diana
1 hour ago
The teacher asked us to try to solve it with Cauchy-Schwarz
– Diana
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
By C-S
$$frac1n+1+frac1n+2+...+frac12n+frac12n+1+frac12n+2+...+frac13n+frac13n+1=$$
$$=left(frac1n+1+frac13n+1right)+left(frac1n+2+frac13nright)+...+left(frac12n+frac12n+2right)+frac12n+1>$$
$$>frac(1+1)^24n+2+frac(1+1)^24n+2+...+frac(1+1)^24n+2+frac12n+1=1.$$
answered 1 hour ago
Michael Rozenberg
91.7k1584181
that's also Titu's lemma, right?
– Diana
1 hour ago
@Diana I don't like this naming. It's Cauchy-Schwarz.
– Michael Rozenberg
1 hour ago
Is the sum 1+1/2+1/3+...+1/n greater than 1/n+1+1/n+2+...+1/3n+1?
– Diana
1 min ago
@Diana Yes, of course!
– Michael Rozenberg
5 secs ago
add a comment |Â
up vote
2
down vote
Here's an alternative answer using $AM-HM$ inequality: $$fraca_1+a_2+cdots+a_nn geq fracnfrac1a_1+frac1a_2+cdots+frac1a_n$$
take $a_1=n+1, a_2=n+2,cdots,a_2n+1=3n+1$, then by above inequality $$frac(n+1)+(n+2)+cdots+(3n+1)2n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ Consequently, $$frac4n^2+4n+12n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ which implies $$frac1n+1+frac1n+2+cdots+frac13n+1 geq frac(2n+1)^24n^2+4n+1=1$$
answered 1 hour ago
Chinnapparaj R
3,385622
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
By C-S
$$frac1n+1+frac1n+2+...+frac12n+frac12n+1+frac12n+2+...+frac13n+frac13n+1=$$
$$=left(frac1n+1+frac13n+1right)+left(frac1n+2+frac13nright)+...+left(frac12n+frac12n+2right)+frac12n+1>$$
$$>frac(1+1)^24n+2+frac(1+1)^24n+2+...+frac(1+1)^24n+2+frac12n+1=1.$$
answered 1 hour ago
Michael Rozenberg
91.7k1584181
that's also Titu's lemma, right?
– Diana
1 hour ago
@Diana I don't like this naming. It's Cauchy-Schwarz.
– Michael Rozenberg
1 hour ago
Is the sum 1+1/2+1/3+...+1/n greater than 1/n+1+1/n+2+...+1/3n+1?
– Diana
1 min ago
@Diana Yes, of course!
– Michael Rozenberg
5 secs ago
add a comment |Â
up vote
4
down vote
accepted
By C-S
$$frac1n+1+frac1n+2+...+frac12n+frac12n+1+frac12n+2+...+frac13n+frac13n+1=$$
$$=left(frac1n+1+frac13n+1right)+left(frac1n+2+frac13nright)+...+left(frac12n+frac12n+2right)+frac12n+1>$$
$$>frac(1+1)^24n+2+frac(1+1)^24n+2+...+frac(1+1)^24n+2+frac12n+1=1.$$
answered 1 hour ago
Michael Rozenberg
91.7k1584181
that's also Titu's lemma, right?
– Diana
1 hour ago
@Diana I don't like this naming. It's Cauchy-Schwarz.
– Michael Rozenberg
1 hour ago
Is the sum 1+1/2+1/3+...+1/n greater than 1/n+1+1/n+2+...+1/3n+1?
– Diana
1 min ago
@Diana Yes, of course!
– Michael Rozenberg
5 secs ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
By C-S
$$frac1n+1+frac1n+2+...+frac12n+frac12n+1+frac12n+2+...+frac13n+frac13n+1=$$
$$=left(frac1n+1+frac13n+1right)+left(frac1n+2+frac13nright)+...+left(frac12n+frac12n+2right)+frac12n+1>$$
$$>frac(1+1)^24n+2+frac(1+1)^24n+2+...+frac(1+1)^24n+2+frac12n+1=1.$$
answered 1 hour ago
Michael Rozenberg
91.7k1584181
By C-S
$$frac1n+1+frac1n+2+...+frac12n+frac12n+1+frac12n+2+...+frac13n+frac13n+1=$$
$$=left(frac1n+1+frac13n+1right)+left(frac1n+2+frac13nright)+...+left(frac12n+frac12n+2right)+frac12n+1>$$
$$>frac(1+1)^24n+2+frac(1+1)^24n+2+...+frac(1+1)^24n+2+frac12n+1=1.$$
answered 1 hour ago
Michael Rozenberg
91.7k1584181
answered 1 hour ago
Michael Rozenberg
91.7k1584181
answered 1 hour ago
Michael Rozenberg
91.7k1584181
answered 1 hour ago
Michael Rozenberg
91.7k1584181
91.7k1584181
that's also Titu's lemma, right?
– Diana
1 hour ago
@Diana I don't like this naming. It's Cauchy-Schwarz.
– Michael Rozenberg
1 hour ago
Is the sum 1+1/2+1/3+...+1/n greater than 1/n+1+1/n+2+...+1/3n+1?
– Diana
1 min ago
@Diana Yes, of course!
– Michael Rozenberg
5 secs ago
add a comment |Â
that's also Titu's lemma, right?
– Diana
1 hour ago
@Diana I don't like this naming. It's Cauchy-Schwarz.
– Michael Rozenberg
1 hour ago
Is the sum 1+1/2+1/3+...+1/n greater than 1/n+1+1/n+2+...+1/3n+1?
– Diana
1 min ago
@Diana Yes, of course!
– Michael Rozenberg
5 secs ago
that's also Titu's lemma, right?
– Diana
1 hour ago
that's also Titu's lemma, right?
– Diana
1 hour ago
@Diana I don't like this naming. It's Cauchy-Schwarz.
– Michael Rozenberg
1 hour ago
@Diana I don't like this naming. It's Cauchy-Schwarz.
– Michael Rozenberg
1 hour ago
Is the sum 1+1/2+1/3+...+1/n greater than 1/n+1+1/n+2+...+1/3n+1?
– Diana
1 min ago
Is the sum 1+1/2+1/3+...+1/n greater than 1/n+1+1/n+2+...+1/3n+1?
– Diana
1 min ago
@Diana Yes, of course!
– Michael Rozenberg
5 secs ago
@Diana Yes, of course!
– Michael Rozenberg
5 secs ago
add a comment |Â
up vote
2
down vote
Here's an alternative answer using $AM-HM$ inequality: $$fraca_1+a_2+cdots+a_nn geq fracnfrac1a_1+frac1a_2+cdots+frac1a_n$$
take $a_1=n+1, a_2=n+2,cdots,a_2n+1=3n+1$, then by above inequality $$frac(n+1)+(n+2)+cdots+(3n+1)2n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ Consequently, $$frac4n^2+4n+12n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ which implies $$frac1n+1+frac1n+2+cdots+frac13n+1 geq frac(2n+1)^24n^2+4n+1=1$$
answered 1 hour ago
Chinnapparaj R
3,385622
add a comment |Â
up vote
2
down vote
Here's an alternative answer using $AM-HM$ inequality: $$fraca_1+a_2+cdots+a_nn geq fracnfrac1a_1+frac1a_2+cdots+frac1a_n$$
take $a_1=n+1, a_2=n+2,cdots,a_2n+1=3n+1$, then by above inequality $$frac(n+1)+(n+2)+cdots+(3n+1)2n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ Consequently, $$frac4n^2+4n+12n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ which implies $$frac1n+1+frac1n+2+cdots+frac13n+1 geq frac(2n+1)^24n^2+4n+1=1$$
answered 1 hour ago
Chinnapparaj R
3,385622
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here's an alternative answer using $AM-HM$ inequality: $$fraca_1+a_2+cdots+a_nn geq fracnfrac1a_1+frac1a_2+cdots+frac1a_n$$
take $a_1=n+1, a_2=n+2,cdots,a_2n+1=3n+1$, then by above inequality $$frac(n+1)+(n+2)+cdots+(3n+1)2n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ Consequently, $$frac4n^2+4n+12n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ which implies $$frac1n+1+frac1n+2+cdots+frac13n+1 geq frac(2n+1)^24n^2+4n+1=1$$
answered 1 hour ago
Chinnapparaj R
3,385622
Here's an alternative answer using $AM-HM$ inequality: $$fraca_1+a_2+cdots+a_nn geq fracnfrac1a_1+frac1a_2+cdots+frac1a_n$$
take $a_1=n+1, a_2=n+2,cdots,a_2n+1=3n+1$, then by above inequality $$frac(n+1)+(n+2)+cdots+(3n+1)2n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ Consequently, $$frac4n^2+4n+12n+1 geq frac2n+1frac1n+1+frac1n+2+cdots+frac13n+1$$ which implies $$frac1n+1+frac1n+2+cdots+frac13n+1 geq frac(2n+1)^24n^2+4n+1=1$$
answered 1 hour ago
Chinnapparaj R
3,385622
edited 1 hour ago
answered 1 hour ago
Chinnapparaj R
3,385622
answered 1 hour ago
Chinnapparaj R
3,385622
answered 1 hour ago
Chinnapparaj R
3,385622
3,385622
add a comment |Â
add a comment |Â
Diana is a new contributor. Be nice, and check out our Code of Conduct.
Diana is a new contributor. Be nice, and check out our Code of Conduct.
Diana is a new contributor. Be nice, and check out our Code of Conduct.
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Are you sure you need Cauchy-Schwarz inequality? It can be solved by simple algebraic manipulation.
– Gibbs
1 hour ago
1
The teacher asked us to try to solve it with Cauchy-Schwarz
– Diana
1 hour ago