# For which $a, b, c$ does the inequality $a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\ge 6abc$ hold?

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

The inequality $a^3+b^3+c^3\ge 3abc$ holds for non-negatives $a, b, c$ but it holds for real numbers as well iff $(a+b+c)\ge 0$ since: $a^2+b^2+c^2-ab-bc-ca\ge 0 \quad \forall a, b, c \in \mathbb{R} \\ \stackrel{a+b+c\ge 0}{\Longleftrightarrow} (a+b+c)(a^2+b^2+c^2-ac-bc-ca)\ge 0 \\ \stackrel{\text{Euler's Formula}}{\Longleftrightarrow} a^3+b^3+c^3-3abc\ge 0 \\ \Leftrightarrow a^3+b^3+c^3\ge 3abc$ (Generally Note that we can prove the same way that $a+b+c\le 0 \Leftrightarrow a^3+b^3+c^3\le 3abc$)

Considering that the inequality $2(a^3+b^3+c^3)\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2$ holds for all real numbers by rearrangement Inequality, I suppose that something similar for the inequality $a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\ge 6abc \quad (*)$ should exist. But unfortunately $(a+b+c)\ge 0$ is not the condition we are looking for...

I don't really know if that helps but $(*) \Leftrightarrow (a+b)(b+c)(c+a)\ge 8abc \Leftrightarrow (a+b+c)(ab+bc+ca)\ge 9abc$ Note by Chris Galanis
4 years, 10 months ago

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