Relative commutativity degree of nonabelian metabelian

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$$fraca^2-b^2c+fracb^2-c^2a+fracc^2+2a^2bgeq frac2ab-2bc+3cab$$

I have tried that :

$ageq bgeq cRightarrow fraca^2-b^2cgeq 0;fracb^2-c^2ageq 0;frac3a^2bgeq frac3acb$

$fraca^2-b^2c+fracb^2-c^2a+fracc^2+2a^2bgeq frac2ab-2bc+3cabLeftrightarrow fracc^2-a^2b+frac3a^2bgeq 2(a-c)+frac3acbLeftrightarrow frac(c-a)(c+a)bgeq 2(a-c)Leftrightarrow c+ageq -2b$ !!??




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edited Dec 8, 2013 at 5:54
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Micah
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$egingroup$
$abc*(LHS-RHS)=-bc^3+ac^3+2abc^2-3a^2c^2+b^3c-2a^2bc+2a^3c-ab^3+a^3b=ab(a^2-b^2)+bc(b^2-c^2)+ac(a-c)(2a-c) ge 0$

it is trivial when$a=b=c$, the = is hold.$implies LHS ge RHS$


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edited Dec 8, 2013 at 6:12
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$egingroup$
Another way to lớn look at it would be:eginalignLHS &= frac(a - b)(a + b)c + frac(b - c)(b + c)a + fracc^2 + 2a^2b \&> (a - b) + (b - c) + fracc^2 + 2a^2b \&= a - c + fracc^2 + 2a^2b. endalignAnd RHS $= 2a - 2c + dfrac3cab$.

eginalign extSo, LHS > RHS &iff fracc^2 + 2a^2 - 3acb > a - c \&iff c^2 + 2a^2 - 3ac > ab - bc \&iff (a - c)^2 + a(a - c) > b(a - c) \&iff a - c + a > b \&iff 2a > b + cendalign

and this last inequality is true since $a > b > c$.


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