For which a,b,ca, b, c does the inequality a2b+ab2+b2c+bc2+c2a+ca2≥6abca^2b+ab^2+b^2c+bc^2+c^2a+ca^2\ge 6abc hold?

I know that it holds for non-negatives a,b,ca, b, c but consider the following:

The inequality a3+b3+c3≥3abca^3+b^3+c^3\ge 3abc holds for non-negatives a,b,ca, b, c but it holds for real numbers as well iff (a+b+c)≥0(a+b+c)\ge 0 since: a2+b2+c2−ab−bc−ca≥0∀a,b,c∈R⟺a+b+c≥0(a+b+c)(a2+b2+c2−ac−bc−ca)≥0⟺Euler’s Formulaa3+b3+c3−3abc≥0⇔a3+b3+c3≥3abca^2+b^2+c^2-ab-bc-ca\ge 0 \quad \forall a, b, c \in \mathbb{R} \\ \stackrel{a+b+c\ge 0}{\Longleftrightarrow} (a+b+c)(a^2+b^2+c^2-ac-bc-ca)\ge 0 \\ \stackrel{\text{Euler's Formula}}{\Longleftrightarrow} a^3+b^3+c^3-3abc\ge 0 \\ \Leftrightarrow a^3+b^3+c^3\ge 3abc (Generally Note that we can prove the same way that a+b+c≤0⇔a3+b3+c3≤3abca+b+c\le 0 \Leftrightarrow a^3+b^3+c^3\le 3abc)

Considering that the inequality 2(a3+b3+c3)≥a2b+ab2+b2c+bc2+c2a+ca22(a^3+b^3+c^3)\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 holds for all real numbers by rearrangement Inequality, I suppose that something similar for the inequality a2b+ab2+b2c+bc2+c2a+ca2≥6abc(∗)a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\ge 6abc \quad (*) should exist. But unfortunately (a+b+c)≥0(a+b+c)\ge 0 is not the condition we are looking for...

I don't really know if that helps but (∗)⇔(a+b)(b+c)(c+a)≥8abc⇔(a+b+c)(ab+bc+ca)≥9abc(*) \Leftrightarrow (a+b)(b+c)(c+a)\ge 8abc \Leftrightarrow (a+b+c)(ab+bc+ca)\ge 9abc

#Algebra

Note by Chris Galanis
4 years, 10 months ago

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