Question: Prove that $(a^3+b^3+c^3)(a^2+b^2+c^2)\ge(a+b+c)(a^2b^2+b^2c^2+c^2a^2)$, where a, b, c > 0.

I did LHS-RHS, broke parentheses and cancelled terms.

I think I might use $a^2+b^2+c^2\ge ab+bc+ca$. Then I only need to prove:

$a^5+b^5+c^5\gt abc(a^2+b^2+c^2)$.

Answer: 
Notice the last inequality is $a^5+b^5+c^5\geq abc(a^2+b^2+c^2)=a^3bc+ab^3c+abc^3$. One way to do it would be to quote Muirhead's Inequality directly, since $(5,0,0)\succ(3,1,1)$. Alternatively, using AM-GM, you obtain:

$$a^5+a^5+a^5+b^5+c^5\geq 5a^3bc$$
$$b^5+b^5+b^5+c^5+a^5\geq 5ab^3c$$
$$c^5+c^5+c^5+a^5+b^5\geq 5abc^3$$

I leave you to finish the proof using the above 3 inequalities if you do not want to use Muirhead's.
Yes, C-S helps here:
$$\sum\_{cyc}a^3\sum\_{cyc}a^2=\frac{\sum\limits\_{cyc}a^3\sum\limits\_{cyc}a\sum\limits\_{cyc}a^2}{\sum\limits\_{cyc}a}\geq\frac{\left(\sum\limits\_{cyc}a^2\right)^2\sum\limits\_{cyc}a^2}{\sum\limits\_{cyc}a}=$$
$$=\frac{\frac{1}{3}\left(\sum\limits\_{cyc}a^2\right)^2\sum\limits\_{cyc}1\sum\limits\_{cyc}a^2}{\sum\limits\_{cyc}a}\geq\frac{\sum\limits\_{cyc}a^2b^2\left(\sum\limits\_{cyc}a\right)^2}{\sum\limits\_{cyc}a}=\sum\limits\_{cyc}a^2b^2\sum\limits\_{cyc}a.$$