Mixing
inequality involving $(xy+yz+zx)^3$ and the $pqr$ method
Clash Royale CLAN TAG#URR8PPP
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For non-negative numbers $x$, $y$, and $z$ the claim is that:
$$(xy+yz+zx)^3+9 x^2 y^2 z^2 geq 4 (x+y+z)(xy+yz+zx)xyz$$
Without loss of generality, one may assume that $x+y+z=1$ so that $xyzlefrac127$ and $xy+yz+zxlefrac13$. I have not seen how Schur's inequality is helpful.
inequality summation substitution symmetric-polynomials
edited yesterday
greedoid
26.1k93373
asked yesterday
Phil
283
add a comment |Â
up vote
0
down vote
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For non-negative numbers $x$, $y$, and $z$ the claim is that:
$$(xy+yz+zx)^3+9 x^2 y^2 z^2 geq 4 (x+y+z)(xy+yz+zx)xyz$$
Without loss of generality, one may assume that $x+y+z=1$ so that $xyzlefrac127$ and $xy+yz+zxlefrac13$. I have not seen how Schur's inequality is helpful.
inequality summation substitution symmetric-polynomials
edited yesterday
greedoid
26.1k93373
asked yesterday
Phil
283
What is exactly your question?
– Oldboy
yesterday
Why you have assumed $x+y+z=1$?
– Sufaid Saleel
yesterday
1
@SufaidSaleel If you expand both sides of the inequality, all terms are of degree 6, so if the claim is true for some triple $(x,y,z)$ then it is also true for $(ax,ay,az)$ for any constant $a$. Therefore it is sufficient to prove the claim in the case $x+y+z=1$.
– alphacapture
yesterday
@alphacapture Got it!
– Sufaid Saleel
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For non-negative numbers $x$, $y$, and $z$ the claim is that:
$$(xy+yz+zx)^3+9 x^2 y^2 z^2 geq 4 (x+y+z)(xy+yz+zx)xyz$$
Without loss of generality, one may assume that $x+y+z=1$ so that $xyzlefrac127$ and $xy+yz+zxlefrac13$. I have not seen how Schur's inequality is helpful.
inequality summation substitution symmetric-polynomials
edited yesterday
greedoid
26.1k93373
asked yesterday
Phil
283
For non-negative numbers $x$, $y$, and $z$ the claim is that:
$$(xy+yz+zx)^3+9 x^2 y^2 z^2 geq 4 (x+y+z)(xy+yz+zx)xyz$$
Without loss of generality, one may assume that $x+y+z=1$ so that $xyzlefrac127$ and $xy+yz+zxlefrac13$. I have not seen how Schur's inequality is helpful.
inequality summation substitution symmetric-polynomials
edited yesterday
greedoid
26.1k93373
asked yesterday
Phil
283
edited yesterday
greedoid
26.1k93373
edited yesterday
greedoid
26.1k93373
edited yesterday
greedoid
26.1k93373
26.1k93373
asked yesterday
Phil
283
asked yesterday
Phil
283
asked yesterday
Phil
283
283
What is exactly your question?
– Oldboy
yesterday
Why you have assumed $x+y+z=1$?
– Sufaid Saleel
yesterday
1
@SufaidSaleel If you expand both sides of the inequality, all terms are of degree 6, so if the claim is true for some triple $(x,y,z)$ then it is also true for $(ax,ay,az)$ for any constant $a$. Therefore it is sufficient to prove the claim in the case $x+y+z=1$.
– alphacapture
yesterday
@alphacapture Got it!
– Sufaid Saleel
yesterday
add a comment |Â
What is exactly your question?
– Oldboy
yesterday
Why you have assumed $x+y+z=1$?
– Sufaid Saleel
yesterday
1
@SufaidSaleel If you expand both sides of the inequality, all terms are of degree 6, so if the claim is true for some triple $(x,y,z)$ then it is also true for $(ax,ay,az)$ for any constant $a$. Therefore it is sufficient to prove the claim in the case $x+y+z=1$.
– alphacapture
yesterday
@alphacapture Got it!
– Sufaid Saleel
yesterday
What is exactly your question?
– Oldboy
yesterday
What is exactly your question?
– Oldboy
yesterday
Why you have assumed $x+y+z=1$?
– Sufaid Saleel
yesterday
Why you have assumed $x+y+z=1$?
– Sufaid Saleel
yesterday
1
1
@SufaidSaleel If you expand both sides of the inequality, all terms are of degree 6, so if the claim is true for some triple $(x,y,z)$ then it is also true for $(ax,ay,az)$ for any constant $a$. Therefore it is sufficient to prove the claim in the case $x+y+z=1$.
– alphacapture
yesterday
@SufaidSaleel If you expand both sides of the inequality, all terms are of degree 6, so if the claim is true for some triple $(x,y,z)$ then it is also true for $(ax,ay,az)$ for any constant $a$. Therefore it is sufficient to prove the claim in the case $x+y+z=1$.
– alphacapture
yesterday
@alphacapture Got it!
– Sufaid Saleel
yesterday
@alphacapture Got it!
– Sufaid Saleel
yesterday
add a comment |Â
1 Answer
1
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1
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Let $xy=c$, $xz=b$ and $yz=a$.
Thus, we need to prove that
$$(a+b+c)^3+9abcgeq4(a+b+c)(ab+ac+bc)$$ or
$$sum_cyc(a^3-a^2b-a^2c+abc)geq0,$$ which is Schur.
answered yesterday
Michael Rozenberg
86.9k1575178
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $xy=c$, $xz=b$ and $yz=a$.
Thus, we need to prove that
$$(a+b+c)^3+9abcgeq4(a+b+c)(ab+ac+bc)$$ or
$$sum_cyc(a^3-a^2b-a^2c+abc)geq0,$$ which is Schur.
answered yesterday
Michael Rozenberg
86.9k1575178
add a comment |Â
up vote
1
down vote
Let $xy=c$, $xz=b$ and $yz=a$.
Thus, we need to prove that
$$(a+b+c)^3+9abcgeq4(a+b+c)(ab+ac+bc)$$ or
$$sum_cyc(a^3-a^2b-a^2c+abc)geq0,$$ which is Schur.
answered yesterday
Michael Rozenberg
86.9k1575178
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $xy=c$, $xz=b$ and $yz=a$.
Thus, we need to prove that
$$(a+b+c)^3+9abcgeq4(a+b+c)(ab+ac+bc)$$ or
$$sum_cyc(a^3-a^2b-a^2c+abc)geq0,$$ which is Schur.
answered yesterday
Michael Rozenberg
86.9k1575178
Let $xy=c$, $xz=b$ and $yz=a$.
Thus, we need to prove that
$$(a+b+c)^3+9abcgeq4(a+b+c)(ab+ac+bc)$$ or
$$sum_cyc(a^3-a^2b-a^2c+abc)geq0,$$ which is Schur.
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
answered yesterday
Michael Rozenberg
86.9k1575178
86.9k1575178
add a comment |Â
add a comment |Â
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What is exactly your question?
– Oldboy
yesterday
Why you have assumed $x+y+z=1$?
– Sufaid Saleel
yesterday
1
@SufaidSaleel If you expand both sides of the inequality, all terms are of degree 6, so if the claim is true for some triple $(x,y,z)$ then it is also true for $(ax,ay,az)$ for any constant $a$. Therefore it is sufficient to prove the claim in the case $x+y+z=1$.
– alphacapture
yesterday
@alphacapture Got it!
– Sufaid Saleel
yesterday