Olympiad Inequality Question I am new to the Olympiad-style questions and I hope someone could correct my proof for this question as I do not have the answer for it. Please leave some constructive criticism if possible so I could improve. Thanks in advance!

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

My attempt:
By AM-GM Inequality,
$$\frac{a^2+b^2}{2}\ge\sqrt{a^2 b^2}=ab$$
$$\frac{a^2+c^2}{2}\ge\sqrt{a^2 c^2} =ac$$
$$\frac{b^2+c^2}{2}\ge\sqrt{b^2 c^2} =bc$$
Next, multiply $a,b,$ or $c$ to get RHS of the inequality wanted above:
$\dfrac{a(a^2+b^2)}{2} \ge a^2b$, $\dfrac{b(b^2+c^2)}{2} \ge b^2c$, $\dfrac{c(a^2+c^2)}{2} \ge ac^2$
Adding up the inequalities gives us:
$$\dfrac{a^3+ab^2+b^3+bc^2+a^2c+c^3}{2} \ge a^2b+b^2c+ac^2$$
and rearranging the inequality gives us:
$$a^3+b^3+c^3 \ge 2(a^2b+b^2c+ac^2)-ab^2-bc^2-a^2c$$ which is generally true.
 A: Hint: $a^3+a^3+b^3 \ge 3a^2b$ by AM-GM. Do it $2$ more times with the pairs $(b,c)$ and $(c,a)$. Then add up the $3$ inequalities, and divide both sides by $3$ to complete the proof. 
A: While your final inequality isn't yet what you want, you can make a small modification to get the the answer.   
The idea here is that the cyclic sums (but in opposite directions) $Y$ and $Z$ are highly related to each other, so if we have several expressions involving them, we can try to cancel out a term (E.g. via Gaussian elimination). 

Let $ X = \sum a^3, Y = \sum a^2b , Z = \sum ab^2$.
You are asked to show that $ X \geq Y$.
You have shown that $ X \geq 2Y - Z$.   
Similarly, we can show that $ X \geq 2Z - Y$ by slightly modifying your step. (Do you see how?)      

 Next, multiply $b,c,$ or $a$ to get: $\frac{b(a^2+b^2)}{2} \geq ab^2$, $\frac{c(b^2+c^2)}{2} \geq bc^2$, $\frac{a(a^2+c^2)}{2} \geq a^2c$.
 Adding up the inequalities gives us: $ X \geq 2Z - Y$.

Then, this gives us $ 3X \geq 2 ( 2Y  - Z) + (2Z - Y) = 3 Y$.
Hence $ X \geq Y$ as desired.  

Moral of the story: Sometimes you're just a stepping stone away.
