Proof of Inequality using AM-GM

One trick that shows up a lot with AM-GM inequalities is to apply AM-GM several times, then combine them all together somehow. Let's try that. AM-GM tells us that: $$ x + y + z \geq 3 \sqrt[3]{xyz} \tag{$\star$} $$ Now for what values of $x,y,z$ would the LHS and RHS of $(\star)$ match up with some of the terms in the LHS and RHS of the desired inequality? In particular, suppose that we want the cube root to be $a^2b$. Then $xyz = a^6b^3$, which suggests that we take $x = y = a^3$ and $z = b^3$. Repeating this, we obtain: \begin{align*} a^3 + a^3 + b^3 &\geq 3 \sqrt[3]{a^3a^3b^3} = 3a^2b \\ b^3 + b^3 + c^3 &\geq 3 \sqrt[3]{b^3b^3c^3} = 3b^2c \\ c^3 + c^3 + a^3 &\geq 3 \sqrt[3]{c^3c^3a^3} = 3c^2a \\ \end{align*} Adding up the above three inequalities and dividing through by $3$ yields the desired inequality. $~~\blacksquare$

Using rearrangement inequality with $a^2,b^2,c^2$ and $a,b,c$ we get $$aa^2+bb^2+cc^2\ge a^2b +b^2c+c^2a$$

You already have a great answer from @Adriano. To use AM-GM here, in general the idea would be to observe exponents on both sides and try finding a convex combination of $(3, 0, 0), (0, 3, 0)$ and $(0, 0, 3)$ which gives you a term like $(2, 1, 0)$.

Muirhead's inequality - if you're familiar with it - assures us this will work as $[3, 0, 0] \succ [2, 1, 0]$

So you may consider the following generic equation with non-negative $\alpha+\beta+\gamma=1$: $$\alpha (3, 0, 0)+\beta (0, 3, 0) +\gamma(0, 0, 3)= (2, 1, 0)$$

Obviously $\alpha = \frac23, \beta = \frac13, \gamma=0$ comes to mind. Thus the basic inequality to use would be the AM-GM: $$\tfrac23a^3 + \tfrac13b^3 \ge a^2b$$

Summing three similar inequalities get you the result.