Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$ 
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.

My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b^2+c^2 &\geq ab+bc+ca\\
2a^2+2b^2+2c^2-2ab-2bc-2ca &\geq 0\\
\left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &\geq 0
\end{align*}
$$
Which is obviously true.

However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:
$$
\begin{align*}
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\
a+b+c &\geq a+b+c
\end{align*}
$$
Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?
Edit: Please don't help me with the original inequality, I'll figure it out.
 A: Just assume, wlog $a\leq b\leq c$. Then this equation is all you need:
$$a^3+b^3+c^3=a^2b+b^2c+c^2a+\underset{\geq 0}{\underbrace{(c^2-a^2)(b-a)}}+\underset{\geq 0}{\underbrace{(c^2-b^2)(c-b)}}\geq a^2b+b^2c+c^2a$$
A: Without making any assumption, just simple $AM\ge GM$
$$a^3+a^3+b^3\ge3a^2b$$
$$b^3+b^3+c^3\ge3b^2c$$
$$c^3+c^3+a^3\ge3c^2a$$
$$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$$
A: (@HaiDangel told me. 
https://diendantoanhoc.net/topic/182934-a3-b3-c3geqq-a2b-b2c-c2a/?p=731023)
A stronger version: Let $a, b, c$ be real numbers with $a + b \ge 0, b + c \ge 0$ and $c+a\ge 0$. Prove that
$$a^3 + b^3 + c^3 \ge a^2b + b^2c + c^2a.$$
I have an SOS expression:
\begin{align}
&a^3 + b^3 + c^3 - a^2b - b^2c - c^2a \\
=\ & \frac{(a^2+b^2-2c^2)^2 + 3(a^2-b^2)^2
+ \sum_{\mathrm{cyc}} 4(a+b)(c+a)(a-b)^2}{8(a+b+c)}.
\end{align}
A: WOLG, Let $c$=Max{$a,b,c$}, then there is 2 cases:
case I: $0<a \le b \le c$, we want to prove  $c^2(c-a) \ge a^2(b-a)+b^2(c-b)$
we have $c^2\ge b^2, c^2\ge a^2 \to $,RHS $\le c^2(b-a)+c^2(c-b)=c^2(c-a)$
case II: $0<b \le a \le c$, we want to prove $a^2(a-b)+c^2(c-a) \ge b^2(c-b)$
we have $a^2\ge b^2,c^2 \ge b^2, \to$LHS $ \ge b^2(a-b)+b^2(c-a)=b^2(c-b)$
to summary 2 cases, we have $a^2(a-b)+b^2(b-c)+c^2(c-a) \ge 0$
QED
