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: 
As the inequality is homogeneous, assume: $a+b+c = 3$.  

We know
$$
(a^2 + b^2 + c^2)^2 \geqslant 3(a^2b^2 + b^2c^2 + c^2a^2) = (a+b+c)(a^2b^2 + b^2c^2 + c^2a^2)
$$
We need to prove now:
$$
a^3 + b^3 + c^3\geqslant a^2 + b^2 + c^2\quad \textrm{with}\quad a+b+c=3
$$
Which is
$$
3(a^3 + b^3 + c^3) \geqslant (a+b+c) (a^2 + b^2 + c^2)
$$Obviously true.  

For proof, after expansion, you can add
$$
a^3 + b^3 + c^3 \geqslant a^2b + b^2c + c^2a\ \textrm{and}\ a^3 + b^3 + c^3 \geqslant ab^2 + bc^2 + ca^2
$$Proof of these are only by AM-GM.
Yes, C-S helps here:
$$\sum\_{cyc}a^3\sum\_{cyc}a^2=\frac{\sum\limits\_{cyc}a^3\sum\limits\_{cyc}a\sum\limits\_{cyc}a^2}{\sum\limits\_{cyc}a}\geq\frac{\left(\sum\limits\_{cyc}a^2\right)^2\sum\limits\_{cyc}a^2}{\sum\limits\_{cyc}a}=$$
$$=\frac{\frac{1}{3}\left(\sum\limits\_{cyc}a^2\right)^2\sum\limits\_{cyc}1\sum\limits\_{cyc}a^2}{\sum\limits\_{cyc}a}\geq\frac{\sum\limits\_{cyc}a^2b^2\left(\sum\limits\_{cyc}a\right)^2}{\sum\limits\_{cyc}a}=\sum\limits\_{cyc}a^2b^2\sum\limits\_{cyc}a.$$