Prove $\sum_{cyc} a(a-b)(a-2b) \ge 0$.Where $a,b,c \ge 0$

Question

I’ve tried splitting this into the sum of two Chinese Dumbass triangles: \begin{gather} \quad0\\ {-}5\quad0\\ \hskip1.5em0\quad\hskip.3em15\hskip.6em{-}5\hskip1em\\ 0\quad{-}5\quad0\hskip1.2em0 \end{gather} And \begin{gather} 1\\ 2\quad2\\ 2\quad{-}15\quad2\\ 1\qquad2\quad2\qquad1 \end{gather} And that fails. Any hint or help would be appreciated!

Answer 1

Let $$ f(a,b,c) := a(a-b)(a-2b)+b(b-c)(b-2c)+c(c-a)(c-2a). $$ The key observation is that the coefficient of $x^2$ in $f(a+x,b+x,c+x)$ is $$ -(a-b)-(b-c)-(c-a) = 0; $$ that is, $f(a+x,b+x,c+x)$ is linear in $x$. Furthermore, the coefficient of $x$ is \begin{multline*} -2b(a-b) - 2c(b-c) - 2a(c-a) \\ = (a^2+b^2-2ab) + (b^2+c^2-2bc) + (c^2+a^2-2ac) \ge 0. \end{multline*} It follows that $f(a+x,b+x,c+x)$ is an increasing function of $x$. Therefore, assuming for definiteness that $\min\{a,b,c\}=a$, we get $$ f(a,b,c) \ge f(0,b-a,c-a). $$ That is, the general case reduces to that where $a=0$, and this special case is very easy to verify.

Answer 2

Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Thus, by AM-GM we obtain: $$\sum_{cyc}a(a-b)(a-2b)=2a(u^2-uv+v^2)+u^3-3u^2v+2uv^2+v^3\geq$$ $$\geq4\left(\frac{u^3}{4}\right)+uv^2+uv^2+v^3-3u^2v\geq7\sqrt[7]{\left(\frac{u^3}{4}\right)^4\left(uv^2\right)^2v^3}-3u^2v=\left(\frac{7}{\sqrt[7]{256}}-3\right)u^2v\geq0.$$