Show that for positive numbers a,b,c,d, $sum_cyc ab leq frac14left(sum_cyc a right)^2$ and … [duplicate]
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Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$
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Let a,b,c,d be four positive real numbers. Show that
$$sum_cyc ab leq frac14left(sum_cyc a right)^2$$
and
$$sum_cyc abc leq frac116left(sum_cyc a right)^3$$
My textbook said these inequalities can be referred from the AM-GM inequality but didn't say how to derive it. I can find only looser bounds.
That is, for each term in the left hand side of the first inequality, by AM-GM inequality,
$$ab leq left(fraca+b2right)^2 leq frac14 left(a+b+c+dright)^2$$
$$Longrightarrow sum_cyc ab leq frac44 left(sum_cyc a right)^2$$
Using the same techniques, we get another loose bound for $sum_cyc abc$.
How could I get the bounds in the original inequalities? Please suggest.
PS. It seems like if these inequalities are true, we may get a general relationship for n positive variables as $sum_cyc a_1 ... a_k leq frac1n^k-1left(sum_cyc a_i right)^k$.
PS2. Sorry for double posting. There was an old thread here: Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$ One answer points to Maclaurin's inequality, but I still cannot figure out how to prove it using AM-GM.
inequality summation symmetric-polynomials a.m.-g.m.-inequality
asked Sep 9 at 12:25
DKSG
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marked as duplicate by Michael Rozenberg inequality
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This question already has an answer here:
Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$
2 answers
Let a,b,c,d be four positive real numbers. Show that
$$sum_cyc ab leq frac14left(sum_cyc a right)^2$$
and
$$sum_cyc abc leq frac116left(sum_cyc a right)^3$$
My textbook said these inequalities can be referred from the AM-GM inequality but didn't say how to derive it. I can find only looser bounds.
That is, for each term in the left hand side of the first inequality, by AM-GM inequality,
$$ab leq left(fraca+b2right)^2 leq frac14 left(a+b+c+dright)^2$$
$$Longrightarrow sum_cyc ab leq frac44 left(sum_cyc a right)^2$$
Using the same techniques, we get another loose bound for $sum_cyc abc$.
How could I get the bounds in the original inequalities? Please suggest.
PS. It seems like if these inequalities are true, we may get a general relationship for n positive variables as $sum_cyc a_1 ... a_k leq frac1n^k-1left(sum_cyc a_i right)^k$.
PS2. Sorry for double posting. There was an old thread here: Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$ One answer points to Maclaurin's inequality, but I still cannot figure out how to prove it using AM-GM.
inequality summation symmetric-polynomials a.m.-g.m.-inequality
asked Sep 9 at 12:25
DKSG
163
marked as duplicate by Michael Rozenberg inequality
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Sep 9 at 13:12
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.
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up vote
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This question already has an answer here:
Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$
2 answers
Let a,b,c,d be four positive real numbers. Show that
$$sum_cyc ab leq frac14left(sum_cyc a right)^2$$
and
$$sum_cyc abc leq frac116left(sum_cyc a right)^3$$
My textbook said these inequalities can be referred from the AM-GM inequality but didn't say how to derive it. I can find only looser bounds.
That is, for each term in the left hand side of the first inequality, by AM-GM inequality,
$$ab leq left(fraca+b2right)^2 leq frac14 left(a+b+c+dright)^2$$
$$Longrightarrow sum_cyc ab leq frac44 left(sum_cyc a right)^2$$
Using the same techniques, we get another loose bound for $sum_cyc abc$.
How could I get the bounds in the original inequalities? Please suggest.
PS. It seems like if these inequalities are true, we may get a general relationship for n positive variables as $sum_cyc a_1 ... a_k leq frac1n^k-1left(sum_cyc a_i right)^k$.
PS2. Sorry for double posting. There was an old thread here: Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$ One answer points to Maclaurin's inequality, but I still cannot figure out how to prove it using AM-GM.
inequality summation symmetric-polynomials a.m.-g.m.-inequality
asked Sep 9 at 12:25
DKSG
163
This question already has an answer here:
Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$
2 answers
Let a,b,c,d be four positive real numbers. Show that
$$sum_cyc ab leq frac14left(sum_cyc a right)^2$$
and
$$sum_cyc abc leq frac116left(sum_cyc a right)^3$$
My textbook said these inequalities can be referred from the AM-GM inequality but didn't say how to derive it. I can find only looser bounds.
That is, for each term in the left hand side of the first inequality, by AM-GM inequality,
$$ab leq left(fraca+b2right)^2 leq frac14 left(a+b+c+dright)^2$$
$$Longrightarrow sum_cyc ab leq frac44 left(sum_cyc a right)^2$$
Using the same techniques, we get another loose bound for $sum_cyc abc$.
How could I get the bounds in the original inequalities? Please suggest.
PS. It seems like if these inequalities are true, we may get a general relationship for n positive variables as $sum_cyc a_1 ... a_k leq frac1n^k-1left(sum_cyc a_i right)^k$.
PS2. Sorry for double posting. There was an old thread here: Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$ One answer points to Maclaurin's inequality, but I still cannot figure out how to prove it using AM-GM.
This question already has an answer here:
Showing $sumlimits_cycablefrac14left(sumlimits_cycaright)^2$
2 answers
inequality summation symmetric-polynomials a.m.-g.m.-inequality
inequality summation symmetric-polynomials a.m.-g.m.-inequality
asked Sep 9 at 12:25
DKSG
163
asked Sep 9 at 12:25
DKSG
163
edited Sep 9 at 14:47
asked Sep 9 at 12:25
DKSG
163
asked Sep 9 at 12:25
DKSG
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asked Sep 9 at 12:25
DKSG
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163
marked as duplicate by Michael Rozenberg inequality
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The first inequality.
By AM-GM
$$sum_cycab=(a+c)(b+d)leqleft(fraca+b+c+d2right)^2=frac14(a+b+c+d)^2.$$
I used that
$$sum_cycab=ab+bc+cd+da=b(a+c)+d(a+c)=(a+c)(b+d).$$
The second inequality follows from Maclaurin.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$ where $v>0$ and $abc+abd+acd+bcd=4w^3$.
Thus, by Maclaurin $$ugeq vgeq w.$$
About Maclaurin see here: https://en.wikipedia.org/wiki/Maclaurin%27s_inequality
Your last inequality is wrong.
Try $n=5$ and $k=2$.
But for $n=5$ and $k=3$ it's true.
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
Can you explain more on the left hand side? There are only 4 terms in your $sum_cyc ab$, but in the questions, there are 6 terms (i.e., ab + bc + ac + cd + ad + bd).
– DKSG
Sep 9 at 12:44
@DKSG I added something. See now.
– Michael Rozenberg
Sep 9 at 12:47
Thank you but I still don't understand why your cyclic sum $sum_cyc ab$ expansion has only 4 terms. It should have 6 terms, shouldn't it?
– DKSG
Sep 9 at 13:00
@DKSG No. It should be four terms. $ab+ac+ad+bc+bd+cd=frac14sumlimits_symab.$ Cyclic it's $arightarrow brightarrow crightarrow drightarrow a$. (Four times "$rightarrow$" give four terms).
– Michael Rozenberg
Sep 9 at 13:02
By the way, there is a very nice proof of the last inequality for $n=5$ and $k=3$. If you want to see a solution open another topic. This topic I'll close because it's duplicate.
– Michael Rozenberg
Sep 9 at 13:09
 |Â
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1 Answer
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1 Answer
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The first inequality.
By AM-GM
$$sum_cycab=(a+c)(b+d)leqleft(fraca+b+c+d2right)^2=frac14(a+b+c+d)^2.$$
I used that
$$sum_cycab=ab+bc+cd+da=b(a+c)+d(a+c)=(a+c)(b+d).$$
The second inequality follows from Maclaurin.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$ where $v>0$ and $abc+abd+acd+bcd=4w^3$.
Thus, by Maclaurin $$ugeq vgeq w.$$
About Maclaurin see here: https://en.wikipedia.org/wiki/Maclaurin%27s_inequality
Your last inequality is wrong.
Try $n=5$ and $k=2$.
But for $n=5$ and $k=3$ it's true.
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
Can you explain more on the left hand side? There are only 4 terms in your $sum_cyc ab$, but in the questions, there are 6 terms (i.e., ab + bc + ac + cd + ad + bd).
– DKSG
Sep 9 at 12:44
@DKSG I added something. See now.
– Michael Rozenberg
Sep 9 at 12:47
Thank you but I still don't understand why your cyclic sum $sum_cyc ab$ expansion has only 4 terms. It should have 6 terms, shouldn't it?
– DKSG
Sep 9 at 13:00
@DKSG No. It should be four terms. $ab+ac+ad+bc+bd+cd=frac14sumlimits_symab.$ Cyclic it's $arightarrow brightarrow crightarrow drightarrow a$. (Four times "$rightarrow$" give four terms).
– Michael Rozenberg
Sep 9 at 13:02
By the way, there is a very nice proof of the last inequality for $n=5$ and $k=3$. If you want to see a solution open another topic. This topic I'll close because it's duplicate.
– Michael Rozenberg
Sep 9 at 13:09
 |Â
show 2 more comments
up vote
1
down vote
The first inequality.
By AM-GM
$$sum_cycab=(a+c)(b+d)leqleft(fraca+b+c+d2right)^2=frac14(a+b+c+d)^2.$$
I used that
$$sum_cycab=ab+bc+cd+da=b(a+c)+d(a+c)=(a+c)(b+d).$$
The second inequality follows from Maclaurin.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$ where $v>0$ and $abc+abd+acd+bcd=4w^3$.
Thus, by Maclaurin $$ugeq vgeq w.$$
About Maclaurin see here: https://en.wikipedia.org/wiki/Maclaurin%27s_inequality
Your last inequality is wrong.
Try $n=5$ and $k=2$.
But for $n=5$ and $k=3$ it's true.
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
Can you explain more on the left hand side? There are only 4 terms in your $sum_cyc ab$, but in the questions, there are 6 terms (i.e., ab + bc + ac + cd + ad + bd).
– DKSG
Sep 9 at 12:44
@DKSG I added something. See now.
– Michael Rozenberg
Sep 9 at 12:47
Thank you but I still don't understand why your cyclic sum $sum_cyc ab$ expansion has only 4 terms. It should have 6 terms, shouldn't it?
– DKSG
Sep 9 at 13:00
@DKSG No. It should be four terms. $ab+ac+ad+bc+bd+cd=frac14sumlimits_symab.$ Cyclic it's $arightarrow brightarrow crightarrow drightarrow a$. (Four times "$rightarrow$" give four terms).
– Michael Rozenberg
Sep 9 at 13:02
By the way, there is a very nice proof of the last inequality for $n=5$ and $k=3$. If you want to see a solution open another topic. This topic I'll close because it's duplicate.
– Michael Rozenberg
Sep 9 at 13:09
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
The first inequality.
By AM-GM
$$sum_cycab=(a+c)(b+d)leqleft(fraca+b+c+d2right)^2=frac14(a+b+c+d)^2.$$
I used that
$$sum_cycab=ab+bc+cd+da=b(a+c)+d(a+c)=(a+c)(b+d).$$
The second inequality follows from Maclaurin.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$ where $v>0$ and $abc+abd+acd+bcd=4w^3$.
Thus, by Maclaurin $$ugeq vgeq w.$$
About Maclaurin see here: https://en.wikipedia.org/wiki/Maclaurin%27s_inequality
Your last inequality is wrong.
Try $n=5$ and $k=2$.
But for $n=5$ and $k=3$ it's true.
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
The first inequality.
By AM-GM
$$sum_cycab=(a+c)(b+d)leqleft(fraca+b+c+d2right)^2=frac14(a+b+c+d)^2.$$
I used that
$$sum_cycab=ab+bc+cd+da=b(a+c)+d(a+c)=(a+c)(b+d).$$
The second inequality follows from Maclaurin.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$ where $v>0$ and $abc+abd+acd+bcd=4w^3$.
Thus, by Maclaurin $$ugeq vgeq w.$$
About Maclaurin see here: https://en.wikipedia.org/wiki/Maclaurin%27s_inequality
Your last inequality is wrong.
Try $n=5$ and $k=2$.
But for $n=5$ and $k=3$ it's true.
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
edited Sep 9 at 12:55
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
answered Sep 9 at 12:42
Michael Rozenberg
89.3k1582179
89.3k1582179
Can you explain more on the left hand side? There are only 4 terms in your $sum_cyc ab$, but in the questions, there are 6 terms (i.e., ab + bc + ac + cd + ad + bd).
– DKSG
Sep 9 at 12:44
@DKSG I added something. See now.
– Michael Rozenberg
Sep 9 at 12:47
Thank you but I still don't understand why your cyclic sum $sum_cyc ab$ expansion has only 4 terms. It should have 6 terms, shouldn't it?
– DKSG
Sep 9 at 13:00
@DKSG No. It should be four terms. $ab+ac+ad+bc+bd+cd=frac14sumlimits_symab.$ Cyclic it's $arightarrow brightarrow crightarrow drightarrow a$. (Four times "$rightarrow$" give four terms).
– Michael Rozenberg
Sep 9 at 13:02
By the way, there is a very nice proof of the last inequality for $n=5$ and $k=3$. If you want to see a solution open another topic. This topic I'll close because it's duplicate.
– Michael Rozenberg
Sep 9 at 13:09
 |Â
show 2 more comments
Can you explain more on the left hand side? There are only 4 terms in your $sum_cyc ab$, but in the questions, there are 6 terms (i.e., ab + bc + ac + cd + ad + bd).
– DKSG
Sep 9 at 12:44
@DKSG I added something. See now.
– Michael Rozenberg
Sep 9 at 12:47
Thank you but I still don't understand why your cyclic sum $sum_cyc ab$ expansion has only 4 terms. It should have 6 terms, shouldn't it?
– DKSG
Sep 9 at 13:00
@DKSG No. It should be four terms. $ab+ac+ad+bc+bd+cd=frac14sumlimits_symab.$ Cyclic it's $arightarrow brightarrow crightarrow drightarrow a$. (Four times "$rightarrow$" give four terms).
– Michael Rozenberg
Sep 9 at 13:02
By the way, there is a very nice proof of the last inequality for $n=5$ and $k=3$. If you want to see a solution open another topic. This topic I'll close because it's duplicate.
– Michael Rozenberg
Sep 9 at 13:09
Can you explain more on the left hand side? There are only 4 terms in your $sum_cyc ab$, but in the questions, there are 6 terms (i.e., ab + bc + ac + cd + ad + bd).
– DKSG
Sep 9 at 12:44
Can you explain more on the left hand side? There are only 4 terms in your $sum_cyc ab$, but in the questions, there are 6 terms (i.e., ab + bc + ac + cd + ad + bd).
– DKSG
Sep 9 at 12:44
@DKSG I added something. See now.
– Michael Rozenberg
Sep 9 at 12:47
@DKSG I added something. See now.
– Michael Rozenberg
Sep 9 at 12:47
Thank you but I still don't understand why your cyclic sum $sum_cyc ab$ expansion has only 4 terms. It should have 6 terms, shouldn't it?
– DKSG
Sep 9 at 13:00
Thank you but I still don't understand why your cyclic sum $sum_cyc ab$ expansion has only 4 terms. It should have 6 terms, shouldn't it?
– DKSG
Sep 9 at 13:00
@DKSG No. It should be four terms. $ab+ac+ad+bc+bd+cd=frac14sumlimits_symab.$ Cyclic it's $arightarrow brightarrow crightarrow drightarrow a$. (Four times "$rightarrow$" give four terms).
– Michael Rozenberg
Sep 9 at 13:02
@DKSG No. It should be four terms. $ab+ac+ad+bc+bd+cd=frac14sumlimits_symab.$ Cyclic it's $arightarrow brightarrow crightarrow drightarrow a$. (Four times "$rightarrow$" give four terms).
– Michael Rozenberg
Sep 9 at 13:02
By the way, there is a very nice proof of the last inequality for $n=5$ and $k=3$. If you want to see a solution open another topic. This topic I'll close because it's duplicate.
– Michael Rozenberg
Sep 9 at 13:09
By the way, there is a very nice proof of the last inequality for $n=5$ and $k=3$. If you want to see a solution open another topic. This topic I'll close because it's duplicate.
– Michael Rozenberg
Sep 9 at 13:09
 |Â
show 2 more comments
