Given positive $a, b, c$, prove that $(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$. 
Given positive $a, b, c$, prove that $$\large (a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$

As a starting point,
$$(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$
$$\iff \frac{a^2 + b^2 + c^2}{ab + bc + ca} \ge \frac{(a^3 + b^3 + c^3)(a + b + c)}{(a^2 + b^2 + c^2)^2}$$
$$\iff \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2(ab + bc + ca)} \ge \frac{ab(a - b)^2 + bc(b - c)^2 + ca(c - a)^2}{(a^2 + b^2 + c^2)^2} \tag 1$$
$$\iff \sum_{cyc}(a - b)^2\left[\frac{1}{2(ab + bc + ca)} - \frac{ab}{(a^2 + b^2 + c^2)^2}\right] \ge 0$$
$$\iff \frac{1}{2} \cdot \sum_{cyc}\frac{[a^4 + b^4 + c^4 + 2(ca)^2 + 2(bc)^2 - 2a^2bc - 2b^2ca](a - b)^2}{(ab + bc + ca)(a^2 + b^2 + c^2)^2} \ge 0$$
$$\iff \frac{\displaystyle (a^4 + b^4 + c^4) \cdot \sum_{cyc}(a - b)^2 + 2\sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2]}{(ab + bc + ca)(a^2 + b^2 + c^2)^2} \ge 0$$
Please note for $(1)$ that $$(a^2 + b^2 + c^2) - (ab + bc + ca) = \frac{(a - b)^2 + (b - c)^2 + (c - a)^2}{2}$$ and $$(a^3 + b^3 + c^3)(a + b + c) - (a^2 + b^2 + c^2)^2 = ab(a - b)^2 + bc(b - c)^2 + ca(c - a)^2$$
Moreover, evaluating $\displaystyle \sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2]$, we have that $$\sum_{cyc}ca(ca - b^2)[(a - b)^2 + (b - c)^2] = \sum_{cyc}ca(ca - b^2)(c^2 + a^2 + 2b^2 - 2bc - 2ab)$$
$$ = \sum_{cyc}[(ca)^2 - b^2ca](c^2 + a^2) + 2bca \cdot \sum_{cyc}(ca - b^2)(b - c - a)$$
$$ = \sum_{cyc}b^2[(ab)^2 - c^2ab + (bc)^2 - a^2bc] - 2bca \cdot \sum_{cyc}b[(ab - c^2) + (bc - a^2) - (ca - b^2)]$$
I'm going to stop here. There was an attempt, one that failed tremendously.

Your SOS  works!
Indeed, by your work we obtain:
$$(a^2+b^2+c^2)^3-(a^3+b^3+c^3)(ab+ac+bc)(a+b+c)=$$
$$=(a^2+b^2+c^2)^2(ab+ac+bc)\left(\frac{a^2+b^2+c^2}{ab+ac+bc}-\frac{(a^3+b^3+c^3)(a+b+c)}{(a^2+b^2+c^2)^2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+b^2+c^2)^2-2ab(ab+ac+bc))\geq$$
$$\geq\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+b^2+c^2)(ab+ac+bc)-2ab(ab+ac+bc))=$$
$$=\frac{1}{2}(ab+ac+bc)\sum_{cyc}(a-b)^2((a-b)^2+c^2)\geq0.$$
Also, we can use $uvw$, which gives a solution immediately.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $a^2+b^2+c^2$, $ab+ac+bc$ and $a+b+c$ dot't depend on $w^3$ and since $$a^3+b^3+c^3=27u^3-27uv^2+3w^3,$$ it's enough to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables.
Since, our inequality is homogeneous, it's enough to assume $b=c=1$, which gives
$$(a-1)^2(a^4-2a+4)\geq0$$ and we are done!
