Dgdrxrt

How to reverse every other sublist of a list?

On the insanity of kings as an argument against Monarchy

"What time...?" or "At what time...?" - what is more grammatically correct?

I looked up a future colleague on linkedin before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?

Falsification in Math vs Science

Is "plugging out" electronic devices an American expression?

How to manage monthly salary

Does it makes sense to buy a new cycle to learn riding?

Can't find the latex code for the ⍎ (down tack jot) symbol

Is the gradient of the self-intersections of a curve zero?

How can I create a character who can assume the widest possible range of creature sizes?

How to make payment on the internet without leaving a money trail?

How to answer pointed "are you quitting" questioning when I don't want them to suspect

Why is Grand Jury testimony secret?

What is the use of option -o in the useradd command?

If Wish Duplicates Simulacrum, Are Existing Duplicates Destroyed?

Feasability of miniature nuclear reactors for humanoid cyborgs

What are the motivations for publishing new editions of an existing textbook, beyond new discoveries in a field?

Is bread bad for ducks?

Does duplicating a spell with wish count as casting that spell?

Geography at the pixel level

Idomatic way to prevent slicing?

I see my dog run

How to deal with fear of taking dependencies

An application of Cauchy-Schwarz inequality for two lists of positive inumbers with equal sum [on hold]

-2

1

$begingroup$

Let $a_1,dots,a_n,b_1,dots,b_n$ be positive numbers with
$$
a_1+dots+a_n=b_1+dots+b_n = m
$$
We need to prove that
$$
frac1m sum_i=1^n fraca_i^2b_ige 1
$$

The hint that I get is to use Cauchy-Schwarz inequality. But I fail to see how this can be applied here.

inequality cauchy-schwarz-inequality

share|cite|improve this question

edited Mar 30 at 10:21

Maria Mazur

50k1361125

asked Mar 30 at 9:35

ablmfablmf

2,58152452

$endgroup$

put on hold as off-topic by Saad, user21820, José Carlos Santos, RRL, GNUSupporter 8964民主女神 地下教會 7 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, user21820, José Carlos Santos, RRL, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

-2

1

$begingroup$

Let $a_1,dots,a_n,b_1,dots,b_n$ be positive numbers with
$$
a_1+dots+a_n=b_1+dots+b_n = m
$$
We need to prove that
$$
frac1m sum_i=1^n fraca_i^2b_ige 1
$$

The hint that I get is to use Cauchy-Schwarz inequality. But I fail to see how this can be applied here.

inequality cauchy-schwarz-inequality

share|cite|improve this question

edited Mar 30 at 10:21

Maria Mazur

50k1361125

asked Mar 30 at 9:35

ablmfablmf

2,58152452

$endgroup$

put on hold as off-topic by Saad, user21820, José Carlos Santos, RRL, GNUSupporter 8964民主女神 地下教會 7 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, user21820, José Carlos Santos, RRL, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

Let $a_1,dots,a_n,b_1,dots,b_n$ be positive numbers with
$$
a_1+dots+a_n=b_1+dots+b_n = m
$$
We need to prove that
$$
frac1m sum_i=1^n fraca_i^2b_ige 1
$$

The hint that I get is to use Cauchy-Schwarz inequality. But I fail to see how this can be applied here.

inequality cauchy-schwarz-inequality

share|cite|improve this question

edited Mar 30 at 10:21

Maria Mazur

50k1361125

asked Mar 30 at 9:35

ablmfablmf

2,58152452

$endgroup$

share|cite|improve this question

edited Mar 30 at 10:21

Maria Mazur

50k1361125

asked Mar 30 at 9:35

ablmfablmf

2,58152452

share|cite|improve this question

edited Mar 30 at 10:21

Maria Mazur

50k1361125

asked Mar 30 at 9:35

ablmfablmf

2,58152452

edited Mar 30 at 10:21

Maria Mazur

50k1361125

edited Mar 30 at 10:21

Maria Mazur

50k1361125

asked Mar 30 at 9:35

ablmfablmf

2,58152452

asked Mar 30 at 9:35

ablmfablmf

2,58152452

4 Answers
4

active

oldest

votes

2

$begingroup$

By Cauchy:

$$(b_1+b_2+...+b_n)cdot(a_1^2over b_1+a_2^2over b_2+...+a_n^2over b_n)geq (a_1+a_2+...+a_n)^2$$

So $$mcdot sum_i=1^nfraca_i^2b_igeq m^2$$

and we are done

share|cite|improve this answer

answered Mar 30 at 9:38

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

2

$begingroup$

Hint: Use Cauchy Schwarz in Engel form
$$frac1msum_i=1^nfraca_i^2b_igeq frac1mfrac(a_1+a_2+...+a_n)^2b_1+b_2+...+b_n$$

share|cite|improve this answer

answered Mar 30 at 9:39

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

add a comment | 

1

$begingroup$

Also, we can use AM-GM:
$$frac1msum_k=1^nfraca_k^2b_k=frac1msum_k=1^nleft(fraca_k^2b_k+b_k-b_kright)geqfrac1msum_k=1^nleft(2sqrtfraca_k^2b_kcdot b_k-b_kright)=frac1msum_k=1^nb_k=1.$$

share|cite|improve this answer

answered Mar 30 at 11:05

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

add a comment | 

1

$begingroup$

You may also use Jensen's inequality using the convexity of $f(x) = frac1x$:

$$frac1msum_i=1^n fraca_i^2b_i = frac1msum_i=1^n a_ifrac1fracb_ia_istackrelJensengeqfrac1frac1msum_i=1^n a_ifracb_ia_i = 1$$

share|cite|improve this answer

answered Mar 30 at 11:43

trancelocationtrancelocation

13.7k1829

$endgroup$

add a comment | 

2

$begingroup$

By Cauchy:

$$(b_1+b_2+...+b_n)cdot(a_1^2over b_1+a_2^2over b_2+...+a_n^2over b_n)geq (a_1+a_2+...+a_n)^2$$

So $$mcdot sum_i=1^nfraca_i^2b_igeq m^2$$

and we are done

share|cite|improve this answer

answered Mar 30 at 9:38

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

2

$begingroup$

By Cauchy:

$$(b_1+b_2+...+b_n)cdot(a_1^2over b_1+a_2^2over b_2+...+a_n^2over b_n)geq (a_1+a_2+...+a_n)^2$$

So $$mcdot sum_i=1^nfraca_i^2b_igeq m^2$$

and we are done

share|cite|improve this answer

answered Mar 30 at 9:38

Maria MazurMaria Mazur

50k1361125

$endgroup$

add a comment | 

$begingroup$

By Cauchy:

$$(b_1+b_2+...+b_n)cdot(a_1^2over b_1+a_2^2over b_2+...+a_n^2over b_n)geq (a_1+a_2+...+a_n)^2$$

So $$mcdot sum_i=1^nfraca_i^2b_igeq m^2$$

and we are done

share|cite|improve this answer

answered Mar 30 at 9:38

Maria MazurMaria Mazur

50k1361125

$endgroup$

share|cite|improve this answer

answered Mar 30 at 9:38

Maria MazurMaria Mazur

50k1361125

answered Mar 30 at 9:38

Maria MazurMaria Mazur

50k1361125

answered Mar 30 at 9:38

Maria MazurMaria Mazur

50k1361125

2

$begingroup$

Hint: Use Cauchy Schwarz in Engel form
$$frac1msum_i=1^nfraca_i^2b_igeq frac1mfrac(a_1+a_2+...+a_n)^2b_1+b_2+...+b_n$$

share|cite|improve this answer

answered Mar 30 at 9:39

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

add a comment | 

2

$begingroup$

Hint: Use Cauchy Schwarz in Engel form
$$frac1msum_i=1^nfraca_i^2b_igeq frac1mfrac(a_1+a_2+...+a_n)^2b_1+b_2+...+b_n$$

share|cite|improve this answer

answered Mar 30 at 9:39

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

add a comment | 

$begingroup$

Hint: Use Cauchy Schwarz in Engel form
$$frac1msum_i=1^nfraca_i^2b_igeq frac1mfrac(a_1+a_2+...+a_n)^2b_1+b_2+...+b_n$$

share|cite|improve this answer

answered Mar 30 at 9:39

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

share|cite|improve this answer

answered Mar 30 at 9:39

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

answered Mar 30 at 9:39

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

answered Mar 30 at 9:39

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

1

$begingroup$

Also, we can use AM-GM:
$$frac1msum_k=1^nfraca_k^2b_k=frac1msum_k=1^nleft(fraca_k^2b_k+b_k-b_kright)geqfrac1msum_k=1^nleft(2sqrtfraca_k^2b_kcdot b_k-b_kright)=frac1msum_k=1^nb_k=1.$$

share|cite|improve this answer

answered Mar 30 at 11:05

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

add a comment | 

1

$begingroup$

Also, we can use AM-GM:
$$frac1msum_k=1^nfraca_k^2b_k=frac1msum_k=1^nleft(fraca_k^2b_k+b_k-b_kright)geqfrac1msum_k=1^nleft(2sqrtfraca_k^2b_kcdot b_k-b_kright)=frac1msum_k=1^nb_k=1.$$

share|cite|improve this answer

answered Mar 30 at 11:05

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

add a comment | 

$begingroup$

Also, we can use AM-GM:
$$frac1msum_k=1^nfraca_k^2b_k=frac1msum_k=1^nleft(fraca_k^2b_k+b_k-b_kright)geqfrac1msum_k=1^nleft(2sqrtfraca_k^2b_kcdot b_k-b_kright)=frac1msum_k=1^nb_k=1.$$

share|cite|improve this answer

answered Mar 30 at 11:05

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

share|cite|improve this answer

answered Mar 30 at 11:05

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 30 at 11:05

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 30 at 11:05

Michael RozenbergMichael Rozenberg

110k1896201

1

$begingroup$

You may also use Jensen's inequality using the convexity of $f(x) = frac1x$:

$$frac1msum_i=1^n fraca_i^2b_i = frac1msum_i=1^n a_ifrac1fracb_ia_istackrelJensengeqfrac1frac1msum_i=1^n a_ifracb_ia_i = 1$$

share|cite|improve this answer

answered Mar 30 at 11:43

trancelocationtrancelocation

13.7k1829

$endgroup$

add a comment | 

1

$begingroup$

You may also use Jensen's inequality using the convexity of $f(x) = frac1x$:

$$frac1msum_i=1^n fraca_i^2b_i = frac1msum_i=1^n a_ifrac1fracb_ia_istackrelJensengeqfrac1frac1msum_i=1^n a_ifracb_ia_i = 1$$

share|cite|improve this answer

answered Mar 30 at 11:43

trancelocationtrancelocation

13.7k1829

$endgroup$

add a comment | 

$begingroup$

You may also use Jensen's inequality using the convexity of $f(x) = frac1x$:

$$frac1msum_i=1^n fraca_i^2b_i = frac1msum_i=1^n a_ifrac1fracb_ia_istackrelJensengeqfrac1frac1msum_i=1^n a_ifracb_ia_i = 1$$

share|cite|improve this answer

answered Mar 30 at 11:43

trancelocationtrancelocation

13.7k1829

$endgroup$

share|cite|improve this answer

answered Mar 30 at 11:43

trancelocationtrancelocation

13.7k1829

answered Mar 30 at 11:43

trancelocationtrancelocation

13.7k1829

answered Mar 30 at 11:43

trancelocationtrancelocation

13.7k1829