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?

Unfortunately, I wasn't able to pick up from where you left off.
Starting from the following inequality:
$$a(b+c-a) \leq b(a+c-b) \leq c(a+b-c) $$
which is true given your initial WLOG $a\geq b\geq c$.
And note that $\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c}$.
You can use Rearrangement Inequality so that:
\begin{align}\frac{1}{a} a(b+c-a) + \frac{1}{b} b(a+c-b) + \frac{1}{c} c(a+b-c) & \geq \frac{1}{c} a(b+c-a) + \frac{1}{a} b(a+c-b) + \frac{1}{b} c(a+b-c) \\ a+b+c &\geq \frac{a(b-a)}{c} + a + \frac{b(c-b)}{a} + b + \frac{c(a-c)}{b} + c \\ 0 &\geq \frac{a(b-a)}{c} + \frac{b(c-b)}{a}  + \frac{c(a-c)}{b} \\ \frac{a(a-b)}{c} + \frac{b(b-c)}{a} + \frac{c(c-a)}{b} & \geq 0 \\ a^2b(a-b) + b^2c(b-c)+c^2a(c-a) &\geq 0\end{align}
Hope it's correct.
