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?
 A: 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
A: For $a,b,c$ sides of a triangle, let $$\begin{matrix}a=y+z\\b=z+x\\c=x+y\end{matrix}$$
Then the inequality becomes $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ This is true by Cauchy's inequality: $$x+y+z\le\sqrt{\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}}\sqrt{y+z+x}$$
A: 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.
A: 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*}
