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?

Method 1:As Chrystomath used Ravi's substituition we are left to prove $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ Now

*

*$x\ge y\ge z\implies x^2\ge y^2\ge z^2,\frac{1}{x}\le \frac{1}{y} \le \frac{1}{z}$ $$x^2\left(\frac{1}{x}\right)+y^2\left(\frac{1}{y}\right)+z^2\left(\frac{1}{z}\right)\le \frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}$$ $$\iff x+y+z\le \frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}$$

*$x\ge z\ge y$ is analogous can you finish it off...


Method 2: As I have already given an answer by rearrangement here is a far more reliable way
WLOG $c=max(a,b,c)$
so let  $a=x,b=x+u,c=x+u+w$ where $x,u,w \ge 0$ also by triangle inequality $w< x$ we have to prove $$-ux^2(x+u)-w{(x+u)}^2(x+u+w)+{(x+u+w)}^2x(u+w)\ge 0$$ $$\iff u^3(x-w)+u^2(x^2-w^2)+w^3x+w^2x^2+u^2x^2+uwx^2\ge 0$$ which is obvious!!
If $a=x+u,b=x,c=x+u+w$ then again the inequality is after expanding $$u^3x+u^2wx+u^2x^2+2uw^2x+w^3z+w^2x^2\ge 0$$ which is true
