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If $a,b,c$ positive real numbers, then I have to prove $ frac 118 sumlimits_cycl^ fraca^2b^2 + sumlimits_cycl^ frac a2a+b+c ge frac 1112$
We have that $frac 118 sumlimits_cycl^ fraca^2b^2 ge frac 318 = frac 16$
If we assume that $x=2a+b+c,y=a+2b+c,z=a+b+2c,$ then $a=frac3x-y-z4,b=frac3y-x-z4,c=frac3z-x-y4,$ and
$sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
Since
$sumlimits_cycl^ (frac xy+frac yx) ge 6$
so we have
$ frac34 ge sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
and I stuck there. Thank you
inequality polynomials a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jul 19 at 13:37
Michael Rozenberg
88.1k1579180
asked Jul 16 at 17:16
Steven
1
add a comment |Â
up vote
1
down vote
favorite
If $a,b,c$ positive real numbers, then I have to prove $ frac 118 sumlimits_cycl^ fraca^2b^2 + sumlimits_cycl^ frac a2a+b+c ge frac 1112$
We have that $frac 118 sumlimits_cycl^ fraca^2b^2 ge frac 318 = frac 16$
If we assume that $x=2a+b+c,y=a+2b+c,z=a+b+2c,$ then $a=frac3x-y-z4,b=frac3y-x-z4,c=frac3z-x-y4,$ and
$sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
Since
$sumlimits_cycl^ (frac xy+frac yx) ge 6$
so we have
$ frac34 ge sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
and I stuck there. Thank you
inequality polynomials a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jul 19 at 13:37
Michael Rozenberg
88.1k1579180
asked Jul 16 at 17:16
Steven
1
Use C-S ans S-S method.
– Michael Rozenberg
Jul 17 at 12:02
What is the S-S method @MichaelRozenberg? Any particular help please?
– Steven
Jul 18 at 11:10
It's the SOS-Schur method.
– Michael Rozenberg
Jul 18 at 21:09
How can I apply it?
– Steven
Jul 18 at 21:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $a,b,c$ positive real numbers, then I have to prove $ frac 118 sumlimits_cycl^ fraca^2b^2 + sumlimits_cycl^ frac a2a+b+c ge frac 1112$
We have that $frac 118 sumlimits_cycl^ fraca^2b^2 ge frac 318 = frac 16$
If we assume that $x=2a+b+c,y=a+2b+c,z=a+b+2c,$ then $a=frac3x-y-z4,b=frac3y-x-z4,c=frac3z-x-y4,$ and
$sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
Since
$sumlimits_cycl^ (frac xy+frac yx) ge 6$
so we have
$ frac34 ge sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
and I stuck there. Thank you
inequality polynomials a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jul 19 at 13:37
Michael Rozenberg
88.1k1579180
asked Jul 16 at 17:16
Steven
1
If $a,b,c$ positive real numbers, then I have to prove $ frac 118 sumlimits_cycl^ fraca^2b^2 + sumlimits_cycl^ frac a2a+b+c ge frac 1112$
We have that $frac 118 sumlimits_cycl^ fraca^2b^2 ge frac 318 = frac 16$
If we assume that $x=2a+b+c,y=a+2b+c,z=a+b+2c,$ then $a=frac3x-y-z4,b=frac3y-x-z4,c=frac3z-x-y4,$ and
$sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
Since
$sumlimits_cycl^ (frac xy+frac yx) ge 6$
so we have
$ frac34 ge sumlimits_cycl^ frac a2a+b+c = frac14 [9 - sumlimits_cycl^ (frac xy+frac yx)]$
and I stuck there. Thank you
inequality polynomials a.m.-g.m.-inequality cauchy-schwarz-inequality
edited Jul 19 at 13:37
Michael Rozenberg
88.1k1579180
asked Jul 16 at 17:16
Steven
1
edited Jul 19 at 13:37
Michael Rozenberg
88.1k1579180
edited Jul 19 at 13:37
Michael Rozenberg
88.1k1579180
edited Jul 19 at 13:37
Michael Rozenberg
88.1k1579180
88.1k1579180
asked Jul 16 at 17:16
Steven
1
asked Jul 16 at 17:16
Steven
1
asked Jul 16 at 17:16
Steven
1
1
Use C-S ans S-S method.
– Michael Rozenberg
Jul 17 at 12:02
What is the S-S method @MichaelRozenberg? Any particular help please?
– Steven
Jul 18 at 11:10
It's the SOS-Schur method.
– Michael Rozenberg
Jul 18 at 21:09
How can I apply it?
– Steven
Jul 18 at 21:52
add a comment |Â
Use C-S ans S-S method.
– Michael Rozenberg
Jul 17 at 12:02
What is the S-S method @MichaelRozenberg? Any particular help please?
– Steven
Jul 18 at 11:10
It's the SOS-Schur method.
– Michael Rozenberg
Jul 18 at 21:09
How can I apply it?
– Steven
Jul 18 at 21:52
Use C-S ans S-S method.
– Michael Rozenberg
Jul 17 at 12:02
Use C-S ans S-S method.
– Michael Rozenberg
Jul 17 at 12:02
What is the S-S method @MichaelRozenberg? Any particular help please?
– Steven
Jul 18 at 11:10
What is the S-S method @MichaelRozenberg? Any particular help please?
– Steven
Jul 18 at 11:10
It's the SOS-Schur method.
– Michael Rozenberg
Jul 18 at 21:09
It's the SOS-Schur method.
– Michael Rozenberg
Jul 18 at 21:09
How can I apply it?
– Steven
Jul 18 at 21:52
How can I apply it?
– Steven
Jul 18 at 21:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $c=maxa,b,c$.
Since by C-S $$sum_cycfraca2a+b+c=sum_cycfraca^22a^2+ab+acgeqfrac(a+b+c)^2sumlimits_cyc(2a^2+ab+ac)=frac(a+b+c)^22sumlimits_cyc(a^2+ab),$$
it's enough to prove that
$$frac118left(fraca^2b^2+fracb^2c^2+fracc^2a^2right)+frac(a+b+c)^22sumlimits_cyc(a^2+ab)geqfrac1112$$ or
$$fraca^2b^2+fracb^2c^2+fracc^2a^2-3geqfrac272-frac9(a+b+c)^2sumlimits_cyc(a^2+ab)$$ or
$$fraca^2b^2+fracb^2a^2-2+fracb^2c^2+fracc^2a^2-fracb^2a^2-1geqfrac9sumlimits_cyc(a^2-ab)2sumlimits_cyc(a^2+ab)$$ or
$$frac(a^2-b^2)^2a^2b^2+frac(c^2-a^2)(c^2-b^2)a^2c^2geqfrac9((a-b)^2+(c-a)(c-b))2sumlimits_cyc(a^2+ab).$$
Id est, it's enough to prove that
$$frac(a+b)^2a^2b^2geqfrac92sumlimits_cyc(a^2+ab)$$ and
$$frac(a+c)(b+c)a^2c^2geqfrac92sumlimits_cyc(a^2+ab).$$
Both these inequalities we can prove by AM-GM.
Indeed, $$2(a+b)^2sum_cyc(a^2+ab)geq2(a+b)^2(a^2+ab+b^2)geq2cdot4cdot3a^2b^2>9a^2b^2 $$ and
$$2(a+c)(b+c)sum_cyc(a^2+ab)geq2(a+c)c(a^2+ac+c^2)geq2c(a+a)cdot3ac=12a^2c^2>9a^2c^2.$$
Done!
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $c=maxa,b,c$.
Since by C-S $$sum_cycfraca2a+b+c=sum_cycfraca^22a^2+ab+acgeqfrac(a+b+c)^2sumlimits_cyc(2a^2+ab+ac)=frac(a+b+c)^22sumlimits_cyc(a^2+ab),$$
it's enough to prove that
$$frac118left(fraca^2b^2+fracb^2c^2+fracc^2a^2right)+frac(a+b+c)^22sumlimits_cyc(a^2+ab)geqfrac1112$$ or
$$fraca^2b^2+fracb^2c^2+fracc^2a^2-3geqfrac272-frac9(a+b+c)^2sumlimits_cyc(a^2+ab)$$ or
$$fraca^2b^2+fracb^2a^2-2+fracb^2c^2+fracc^2a^2-fracb^2a^2-1geqfrac9sumlimits_cyc(a^2-ab)2sumlimits_cyc(a^2+ab)$$ or
$$frac(a^2-b^2)^2a^2b^2+frac(c^2-a^2)(c^2-b^2)a^2c^2geqfrac9((a-b)^2+(c-a)(c-b))2sumlimits_cyc(a^2+ab).$$
Id est, it's enough to prove that
$$frac(a+b)^2a^2b^2geqfrac92sumlimits_cyc(a^2+ab)$$ and
$$frac(a+c)(b+c)a^2c^2geqfrac92sumlimits_cyc(a^2+ab).$$
Both these inequalities we can prove by AM-GM.
Indeed, $$2(a+b)^2sum_cyc(a^2+ab)geq2(a+b)^2(a^2+ab+b^2)geq2cdot4cdot3a^2b^2>9a^2b^2 $$ and
$$2(a+c)(b+c)sum_cyc(a^2+ab)geq2(a+c)c(a^2+ac+c^2)geq2c(a+a)cdot3ac=12a^2c^2>9a^2c^2.$$
Done!
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
add a comment |Â
up vote
2
down vote
accepted
Let $c=maxa,b,c$.
Since by C-S $$sum_cycfraca2a+b+c=sum_cycfraca^22a^2+ab+acgeqfrac(a+b+c)^2sumlimits_cyc(2a^2+ab+ac)=frac(a+b+c)^22sumlimits_cyc(a^2+ab),$$
it's enough to prove that
$$frac118left(fraca^2b^2+fracb^2c^2+fracc^2a^2right)+frac(a+b+c)^22sumlimits_cyc(a^2+ab)geqfrac1112$$ or
$$fraca^2b^2+fracb^2c^2+fracc^2a^2-3geqfrac272-frac9(a+b+c)^2sumlimits_cyc(a^2+ab)$$ or
$$fraca^2b^2+fracb^2a^2-2+fracb^2c^2+fracc^2a^2-fracb^2a^2-1geqfrac9sumlimits_cyc(a^2-ab)2sumlimits_cyc(a^2+ab)$$ or
$$frac(a^2-b^2)^2a^2b^2+frac(c^2-a^2)(c^2-b^2)a^2c^2geqfrac9((a-b)^2+(c-a)(c-b))2sumlimits_cyc(a^2+ab).$$
Id est, it's enough to prove that
$$frac(a+b)^2a^2b^2geqfrac92sumlimits_cyc(a^2+ab)$$ and
$$frac(a+c)(b+c)a^2c^2geqfrac92sumlimits_cyc(a^2+ab).$$
Both these inequalities we can prove by AM-GM.
Indeed, $$2(a+b)^2sum_cyc(a^2+ab)geq2(a+b)^2(a^2+ab+b^2)geq2cdot4cdot3a^2b^2>9a^2b^2 $$ and
$$2(a+c)(b+c)sum_cyc(a^2+ab)geq2(a+c)c(a^2+ac+c^2)geq2c(a+a)cdot3ac=12a^2c^2>9a^2c^2.$$
Done!
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $c=maxa,b,c$.
Since by C-S $$sum_cycfraca2a+b+c=sum_cycfraca^22a^2+ab+acgeqfrac(a+b+c)^2sumlimits_cyc(2a^2+ab+ac)=frac(a+b+c)^22sumlimits_cyc(a^2+ab),$$
it's enough to prove that
$$frac118left(fraca^2b^2+fracb^2c^2+fracc^2a^2right)+frac(a+b+c)^22sumlimits_cyc(a^2+ab)geqfrac1112$$ or
$$fraca^2b^2+fracb^2c^2+fracc^2a^2-3geqfrac272-frac9(a+b+c)^2sumlimits_cyc(a^2+ab)$$ or
$$fraca^2b^2+fracb^2a^2-2+fracb^2c^2+fracc^2a^2-fracb^2a^2-1geqfrac9sumlimits_cyc(a^2-ab)2sumlimits_cyc(a^2+ab)$$ or
$$frac(a^2-b^2)^2a^2b^2+frac(c^2-a^2)(c^2-b^2)a^2c^2geqfrac9((a-b)^2+(c-a)(c-b))2sumlimits_cyc(a^2+ab).$$
Id est, it's enough to prove that
$$frac(a+b)^2a^2b^2geqfrac92sumlimits_cyc(a^2+ab)$$ and
$$frac(a+c)(b+c)a^2c^2geqfrac92sumlimits_cyc(a^2+ab).$$
Both these inequalities we can prove by AM-GM.
Indeed, $$2(a+b)^2sum_cyc(a^2+ab)geq2(a+b)^2(a^2+ab+b^2)geq2cdot4cdot3a^2b^2>9a^2b^2 $$ and
$$2(a+c)(b+c)sum_cyc(a^2+ab)geq2(a+c)c(a^2+ac+c^2)geq2c(a+a)cdot3ac=12a^2c^2>9a^2c^2.$$
Done!
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
Let $c=maxa,b,c$.
Since by C-S $$sum_cycfraca2a+b+c=sum_cycfraca^22a^2+ab+acgeqfrac(a+b+c)^2sumlimits_cyc(2a^2+ab+ac)=frac(a+b+c)^22sumlimits_cyc(a^2+ab),$$
it's enough to prove that
$$frac118left(fraca^2b^2+fracb^2c^2+fracc^2a^2right)+frac(a+b+c)^22sumlimits_cyc(a^2+ab)geqfrac1112$$ or
$$fraca^2b^2+fracb^2c^2+fracc^2a^2-3geqfrac272-frac9(a+b+c)^2sumlimits_cyc(a^2+ab)$$ or
$$fraca^2b^2+fracb^2a^2-2+fracb^2c^2+fracc^2a^2-fracb^2a^2-1geqfrac9sumlimits_cyc(a^2-ab)2sumlimits_cyc(a^2+ab)$$ or
$$frac(a^2-b^2)^2a^2b^2+frac(c^2-a^2)(c^2-b^2)a^2c^2geqfrac9((a-b)^2+(c-a)(c-b))2sumlimits_cyc(a^2+ab).$$
Id est, it's enough to prove that
$$frac(a+b)^2a^2b^2geqfrac92sumlimits_cyc(a^2+ab)$$ and
$$frac(a+c)(b+c)a^2c^2geqfrac92sumlimits_cyc(a^2+ab).$$
Both these inequalities we can prove by AM-GM.
Indeed, $$2(a+b)^2sum_cyc(a^2+ab)geq2(a+b)^2(a^2+ab+b^2)geq2cdot4cdot3a^2b^2>9a^2b^2 $$ and
$$2(a+c)(b+c)sum_cyc(a^2+ab)geq2(a+c)c(a^2+ac+c^2)geq2c(a+a)cdot3ac=12a^2c^2>9a^2c^2.$$
Done!
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
answered Jul 19 at 13:35
Michael Rozenberg
88.1k1579180
88.1k1579180
add a comment |Â
add a comment |Â
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Use C-S ans S-S method.
– Michael Rozenberg
Jul 17 at 12:02
What is the S-S method @MichaelRozenberg? Any particular help please?
– Steven
Jul 18 at 11:10
It's the SOS-Schur method.
– Michael Rozenberg
Jul 18 at 21:09
How can I apply it?
– Steven
Jul 18 at 21:52