Let $a,b$ and $c$ be the sides of a triangle. Prove that
$3a^2b + 3b^2c + 3c^2a - 3abc -2b^2a - 2c^2b - 2a^2c \ge0$
I tried to use $a = x+y, b = x + z$ and $c = y+z$ substitutions but it didn't work.Also AM-GM and Cauchy-Schwarz inequalities were not useful.
It works!
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, we need to prove that $$\sum_{cyc}\left(\frac{3(y+z)}{x+y}-\frac{2(y+z)}{x+z}-1\right)\geq0$$ or $$\sum_{cyc}\frac{(y+z)(3x+3z-2x-2y)}{(x+y)(x+z)}\geq3$$ or $$\sum_{cyc}(y+z)^2(x-2y+3z)\geq3\prod_{cyc}(x+y)$$ or
$$\sum_{cyc}(y^2+2yz+z^2)(x-2y+3z)\geq3\prod_{cyc}(x+y)$$ or $$\sum_{cyc}(x^2z-2x^3+3x^2y+2xyz-4x^2y+6x^2z+x^2y-2x^2z+3x^3)\geq$$ $$\geq\sum_{cyc}(3x^2y+3x^2z+2xyz)$$ or $$\sum_{cyc}(x^3-3x^2y+2x^2z)\geq0.$$
Now, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus, $u\geq0$ and $v\geq0$ and we need to prove that $$2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3\geq0,$$ which is true because $$u^2-uv+v^2=\left(u-\frac{v}{2}\right)^2+\frac{3}{4}v^2\geq0$$ and by AM-GM $$u^3-3u^2v+2uv^2+v^3=5\cdot\frac{u^3}{5}+3\cdot\frac{2uv^2}{3}+v^3-3u^2v\geq$$ $$\geq9\sqrt[9]{\left(\frac{u^3}{5}\right)^5\left(\frac{2uv^2}{3}\right)^3v^3}-3u^2v=\left(9\sqrt[9]{\left(\frac{1}{5}\right)^5\left(\frac{2}{3}\right)^3}-3\right)u^2v\geq0.$$