Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.
There are $2$ cases to consider:
Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:
\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\ 
& = abc(a-b) + abc(b-c) +  abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}
Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?

The standard substitution is $x=b+c-a , y=c+a-b , z=a+b-c $ ($x,y,z \geq 0$ by the triangle inequality.) These invert to give
\begin{eqnarray*}
2a&=&y+z \\ 2b&=&z+x \\ 2c&=&x+y.
\end{eqnarray*}
Subbing these gives
\begin{eqnarray*}
x^3z+y^3x+z^3x-xyz(x+y+z) \geq 0
\end{eqnarray*}
which you could $\color{red}{\text{not}}$ say is true by Muirhead's inequality ... but can be shown to be true because
\begin{eqnarray*}
xy(y-z)^2+yz(z-x)^2+zx(x-y)^2 \geq 0.
\end{eqnarray*}
