Show that $frac {a_1^2}{a_2}+frac {a_2^2}{a_3}+…+frac {a_n^2}{a_1}geq a_1+a_2+…+a_n$ using AM-GM....
$begingroup$
Given $a_1,a_2,...,a_n$ be positive reals. Show that $displaystylefrac {a_1^2}{a_2}+frac {a_2^2}{a_3}+...+frac {a_n^2}{a_1}geq a_1+a_2+...+a_n$ using AM-GM.
I know how to slve it using rearrangement inequality, but I can't. How should I apply AM-GM? Thanks.
inequality
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
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closed as off-topic by Carl Mummert, Brahadeesh, user21820, amWhy, Did Dec 11 '18 at 19:22
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." – Carl Mummert, Brahadeesh, user21820, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given $a_1,a_2,...,a_n$ be positive reals. Show that $displaystylefrac {a_1^2}{a_2}+frac {a_2^2}{a_3}+...+frac {a_n^2}{a_1}geq a_1+a_2+...+a_n$ using AM-GM.
I know how to slve it using rearrangement inequality, but I can't. How should I apply AM-GM? Thanks.
inequality
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
$endgroup$
closed as off-topic by Carl Mummert, Brahadeesh, user21820, amWhy, Did Dec 11 '18 at 19:22
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." – Carl Mummert, Brahadeesh, user21820, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Given $a_1,a_2,...,a_n$ be positive reals. Show that $displaystylefrac {a_1^2}{a_2}+frac {a_2^2}{a_3}+...+frac {a_n^2}{a_1}geq a_1+a_2+...+a_n$ using AM-GM.
I know how to slve it using rearrangement inequality, but I can't. How should I apply AM-GM? Thanks.
inequality
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
$endgroup$
Given $a_1,a_2,...,a_n$ be positive reals. Show that $displaystylefrac {a_1^2}{a_2}+frac {a_2^2}{a_3}+...+frac {a_n^2}{a_1}geq a_1+a_2+...+a_n$ using AM-GM.
I know how to slve it using rearrangement inequality, but I can't. How should I apply AM-GM? Thanks.
inequality
inequality
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
asked May 16 '13 at 12:44
A. ChuA. Chu
6,99083283
6,99083283
closed as off-topic by Carl Mummert, Brahadeesh, user21820, amWhy, Did Dec 11 '18 at 19:22
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." – Carl Mummert, Brahadeesh, user21820, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Carl Mummert, Brahadeesh, user21820, amWhy, Did Dec 11 '18 at 19:22
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." – Carl Mummert, Brahadeesh, user21820, amWhy, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$dfrac{a^2_{1}}{a_{2}}+a_{2}ge 2a_{1}$$
$$dfrac{a^2_{2}}{a_{3}}+a_{3}ge 2a_{2}$$
$$cdotscdots$$
$$dfrac{a^2_{n}}{a_{1}}+a_{1}ge 2a_{n}$$
add all inequalities and you're done!
edited Jul 10 '13 at 15:47
dajoker
1,884920
answered May 16 '13 at 12:56
math110math110
32.5k457217
$endgroup$
$begingroup$
and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you
$endgroup$
– math110
May 16 '13 at 12:56
1
$begingroup$
use this cauchy-Schwarz $$left(dfrac{a^2_{1}}{a_{2}}+dfrac{a^2_{2}}{a_{3}}+cdots+dfrac{a^2_{n}}{a_{1}}right)(a_{2}+a_{3}+cdots+a_{n}+a_{1})ge (a_{1}+a_{2}+cdots+a_{n})^2$$
$endgroup$
– math110
May 16 '13 at 13:00
$begingroup$
Nice. What do you mean by "and all equality"?
$endgroup$
– Julien
May 16 '13 at 13:32
$begingroup$
Thank you, I mean add,@julien
$endgroup$
– math110
May 16 '13 at 13:45
add a comment |
$begingroup$
other nice methods
$$Longleftrightarrow left(dfrac{a^2_{1}}{a_{2}}-2a_{1}+a_{2}right)+left(dfrac{a^2_{2}}{a_{3}}-2a_{2}+a_{3}right)+cdots+left(dfrac{a^2_{n}}{a_{1}}-2a_{n}+a_{1}right)ge 0$$
$$Longleftrightarrow left(dfrac{a_{1}}{sqrt{a_{2}}}-sqrt{a_{2}}right)^2+cdots+left(dfrac{a_{n}}{sqrt{a_{1}}}-sqrt{a_{1}}right)^2ge 0$$
answered May 16 '13 at 13:09
math110math110
32.5k457217
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$dfrac{a^2_{1}}{a_{2}}+a_{2}ge 2a_{1}$$
$$dfrac{a^2_{2}}{a_{3}}+a_{3}ge 2a_{2}$$
$$cdotscdots$$
$$dfrac{a^2_{n}}{a_{1}}+a_{1}ge 2a_{n}$$
add all inequalities and you're done!
edited Jul 10 '13 at 15:47
dajoker
1,884920
answered May 16 '13 at 12:56
math110math110
32.5k457217
$endgroup$
$begingroup$
and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you
$endgroup$
– math110
May 16 '13 at 12:56
1
$begingroup$
use this cauchy-Schwarz $$left(dfrac{a^2_{1}}{a_{2}}+dfrac{a^2_{2}}{a_{3}}+cdots+dfrac{a^2_{n}}{a_{1}}right)(a_{2}+a_{3}+cdots+a_{n}+a_{1})ge (a_{1}+a_{2}+cdots+a_{n})^2$$
$endgroup$
– math110
May 16 '13 at 13:00
$begingroup$
Nice. What do you mean by "and all equality"?
$endgroup$
– Julien
May 16 '13 at 13:32
$begingroup$
Thank you, I mean add,@julien
$endgroup$
– math110
May 16 '13 at 13:45
add a comment |
$begingroup$
$$dfrac{a^2_{1}}{a_{2}}+a_{2}ge 2a_{1}$$
$$dfrac{a^2_{2}}{a_{3}}+a_{3}ge 2a_{2}$$
$$cdotscdots$$
$$dfrac{a^2_{n}}{a_{1}}+a_{1}ge 2a_{n}$$
add all inequalities and you're done!
edited Jul 10 '13 at 15:47
dajoker
1,884920
answered May 16 '13 at 12:56
math110math110
32.5k457217
$endgroup$
$begingroup$
and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you
$endgroup$
– math110
May 16 '13 at 12:56
1
$begingroup$
use this cauchy-Schwarz $$left(dfrac{a^2_{1}}{a_{2}}+dfrac{a^2_{2}}{a_{3}}+cdots+dfrac{a^2_{n}}{a_{1}}right)(a_{2}+a_{3}+cdots+a_{n}+a_{1})ge (a_{1}+a_{2}+cdots+a_{n})^2$$
$endgroup$
– math110
May 16 '13 at 13:00
$begingroup$
Nice. What do you mean by "and all equality"?
$endgroup$
– Julien
May 16 '13 at 13:32
$begingroup$
Thank you, I mean add,@julien
$endgroup$
– math110
May 16 '13 at 13:45
add a comment |
$begingroup$
$$dfrac{a^2_{1}}{a_{2}}+a_{2}ge 2a_{1}$$
$$dfrac{a^2_{2}}{a_{3}}+a_{3}ge 2a_{2}$$
$$cdotscdots$$
$$dfrac{a^2_{n}}{a_{1}}+a_{1}ge 2a_{n}$$
add all inequalities and you're done!
edited Jul 10 '13 at 15:47
dajoker
1,884920
answered May 16 '13 at 12:56
math110math110
32.5k457217
$endgroup$
$$dfrac{a^2_{1}}{a_{2}}+a_{2}ge 2a_{1}$$
$$dfrac{a^2_{2}}{a_{3}}+a_{3}ge 2a_{2}$$
$$cdotscdots$$
$$dfrac{a^2_{n}}{a_{1}}+a_{1}ge 2a_{n}$$
add all inequalities and you're done!
edited Jul 10 '13 at 15:47
dajoker
1,884920
answered May 16 '13 at 12:56
math110math110
32.5k457217
edited Jul 10 '13 at 15:47
dajoker
1,884920
edited Jul 10 '13 at 15:47
dajoker
1,884920
edited Jul 10 '13 at 15:47
dajoker
1,884920
1,884920
answered May 16 '13 at 12:56
math110math110
32.5k457217
answered May 16 '13 at 12:56
math110math110
32.5k457217
answered May 16 '13 at 12:56
math110math110
32.5k457217
32.5k457217
$begingroup$
and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you
$endgroup$
– math110
May 16 '13 at 12:56
1
$begingroup$
use this cauchy-Schwarz $$left(dfrac{a^2_{1}}{a_{2}}+dfrac{a^2_{2}}{a_{3}}+cdots+dfrac{a^2_{n}}{a_{1}}right)(a_{2}+a_{3}+cdots+a_{n}+a_{1})ge (a_{1}+a_{2}+cdots+a_{n})^2$$
$endgroup$
– math110
May 16 '13 at 13:00
$begingroup$
Nice. What do you mean by "and all equality"?
$endgroup$
– Julien
May 16 '13 at 13:32
$begingroup$
Thank you, I mean add,@julien
$endgroup$
– math110
May 16 '13 at 13:45
add a comment |
$begingroup$
and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you
$endgroup$
– math110
May 16 '13 at 12:56
1
$begingroup$
use this cauchy-Schwarz $$left(dfrac{a^2_{1}}{a_{2}}+dfrac{a^2_{2}}{a_{3}}+cdots+dfrac{a^2_{n}}{a_{1}}right)(a_{2}+a_{3}+cdots+a_{n}+a_{1})ge (a_{1}+a_{2}+cdots+a_{n})^2$$
$endgroup$
– math110
May 16 '13 at 13:00
$begingroup$
Nice. What do you mean by "and all equality"?
$endgroup$
– Julien
May 16 '13 at 13:32
$begingroup$
Thank you, I mean add,@julien
$endgroup$
– math110
May 16 '13 at 13:45
$begingroup$
and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you
$endgroup$
– math110
May 16 '13 at 12:56
$begingroup$
and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you
$endgroup$
– math110
May 16 '13 at 12:56
1
1
$begingroup$
use this cauchy-Schwarz $$left(dfrac{a^2_{1}}{a_{2}}+dfrac{a^2_{2}}{a_{3}}+cdots+dfrac{a^2_{n}}{a_{1}}right)(a_{2}+a_{3}+cdots+a_{n}+a_{1})ge (a_{1}+a_{2}+cdots+a_{n})^2$$
$endgroup$
– math110
May 16 '13 at 13:00
$begingroup$
use this cauchy-Schwarz $$left(dfrac{a^2_{1}}{a_{2}}+dfrac{a^2_{2}}{a_{3}}+cdots+dfrac{a^2_{n}}{a_{1}}right)(a_{2}+a_{3}+cdots+a_{n}+a_{1})ge (a_{1}+a_{2}+cdots+a_{n})^2$$
$endgroup$
– math110
May 16 '13 at 13:00
$begingroup$
Nice. What do you mean by "and all equality"?
$endgroup$
– Julien
May 16 '13 at 13:32
$begingroup$
Nice. What do you mean by "and all equality"?
$endgroup$
– Julien
May 16 '13 at 13:32
$begingroup$
Thank you, I mean add,@julien
$endgroup$
– math110
May 16 '13 at 13:45
$begingroup$
Thank you, I mean add,@julien
$endgroup$
– math110
May 16 '13 at 13:45
add a comment |
$begingroup$
other nice methods
$$Longleftrightarrow left(dfrac{a^2_{1}}{a_{2}}-2a_{1}+a_{2}right)+left(dfrac{a^2_{2}}{a_{3}}-2a_{2}+a_{3}right)+cdots+left(dfrac{a^2_{n}}{a_{1}}-2a_{n}+a_{1}right)ge 0$$
$$Longleftrightarrow left(dfrac{a_{1}}{sqrt{a_{2}}}-sqrt{a_{2}}right)^2+cdots+left(dfrac{a_{n}}{sqrt{a_{1}}}-sqrt{a_{1}}right)^2ge 0$$
answered May 16 '13 at 13:09
math110math110
32.5k457217
$endgroup$
add a comment |
$begingroup$
other nice methods
$$Longleftrightarrow left(dfrac{a^2_{1}}{a_{2}}-2a_{1}+a_{2}right)+left(dfrac{a^2_{2}}{a_{3}}-2a_{2}+a_{3}right)+cdots+left(dfrac{a^2_{n}}{a_{1}}-2a_{n}+a_{1}right)ge 0$$
$$Longleftrightarrow left(dfrac{a_{1}}{sqrt{a_{2}}}-sqrt{a_{2}}right)^2+cdots+left(dfrac{a_{n}}{sqrt{a_{1}}}-sqrt{a_{1}}right)^2ge 0$$
answered May 16 '13 at 13:09
math110math110
32.5k457217
$endgroup$
add a comment |
$begingroup$
other nice methods
$$Longleftrightarrow left(dfrac{a^2_{1}}{a_{2}}-2a_{1}+a_{2}right)+left(dfrac{a^2_{2}}{a_{3}}-2a_{2}+a_{3}right)+cdots+left(dfrac{a^2_{n}}{a_{1}}-2a_{n}+a_{1}right)ge 0$$
$$Longleftrightarrow left(dfrac{a_{1}}{sqrt{a_{2}}}-sqrt{a_{2}}right)^2+cdots+left(dfrac{a_{n}}{sqrt{a_{1}}}-sqrt{a_{1}}right)^2ge 0$$
answered May 16 '13 at 13:09
math110math110
32.5k457217
$endgroup$
other nice methods
$$Longleftrightarrow left(dfrac{a^2_{1}}{a_{2}}-2a_{1}+a_{2}right)+left(dfrac{a^2_{2}}{a_{3}}-2a_{2}+a_{3}right)+cdots+left(dfrac{a^2_{n}}{a_{1}}-2a_{n}+a_{1}right)ge 0$$
$$Longleftrightarrow left(dfrac{a_{1}}{sqrt{a_{2}}}-sqrt{a_{2}}right)^2+cdots+left(dfrac{a_{n}}{sqrt{a_{1}}}-sqrt{a_{1}}right)^2ge 0$$
answered May 16 '13 at 13:09
math110math110
32.5k457217
answered May 16 '13 at 13:09
math110math110
32.5k457217
answered May 16 '13 at 13:09
math110math110
32.5k457217
answered May 16 '13 at 13:09
math110math110
32.5k457217
32.5k457217
add a comment |
add a comment |
