70. Proof: (1) When $a \geqslant b \geqslant c$, $a^{3} b+b^{3} c+c^{3} a-a b^{3}-b c^{3}-c a^{3}=(a+$ $b+c) \cdot(a-b)(b-c)(a-c) \geqslant 0$. Therefore, it suffices to prove that inequality (1) holds when $a \geqslant b \geqslant c$.

Since
$$\begin{array}{l} 2 \sum b c \cdot \sum(b-c)^{2}+3 a b c \sum a-3 \sum a^{3} b= \\ 2 \sum b c \cdot \sum(b-c)^{2}-3 \sum a b(a-c)^{2}= \\ (2 b c+2 c a-a b)(a-c)^{2}+(2 c a+2 a b-b c)(a-b)^{2}+ \\ (2 a b+2 b c-c a)(b-c)^{2}= \\ (4 a c+b c+a b)(a-b)^{2}+(4 b c+a c+a b)(b-c)^{2}+ \\ 2(2 b c+2 a c-a b)(a-b)(b-c)= \\ 4 a c\left[(a-b)^{2}+(a-b)(b-c)\right]+4 b c\left[(b-c)^{2}+(a-b)(b-c)\right]+ \\ a b\left[(a-b)^{2}+(b-c)^{2}-2(a-b)(b-c)\right]+b c(a-b)^{2}+a c(b-c)^{2} \geqslant \end{array}$$
$$0$$

This shows that inequality (1) holds.
(2) Since $\sum a \cdot\left(a^{2} b+b^{2} c+c^{2} a+a b c\right)=\sum a^{3} b+\left(\sum b c\right)^{2}$, we have
$$\begin{array}{l} \sum a \cdot\left(a^{2} b+b^{2} c+c^{2} a+a b c\right)-\left(\sum b c\right)^{2} \leqslant \\ a b c \cdot \sum a+\frac{2}{3} \sum b c \cdot \sum(b-c)^{2} \end{array}$$

Applying inequality (1), we get
$$\begin{array}{c} \sum a^{2} b \leqslant \frac{\left(\sum b c\right)^{2}+\frac{2}{3} \sum b c \cdot \sum(b-c)^{2}}{\sum a}= \\ \frac{\sum b c \cdot\left[4\left(\sum a\right)^{2}-9 \sum b c\right]}{3 \sum a} \end{array}$$

Note: The above inequality (1) and problem 29: $\sum a^{3} b+\left(\sum b c\right)^{2} \leqslant \frac{4}{27}\left(\sum a\right)^{4}$, as well as the inequality in Example 15 of Chapter 2 "Incremental Comparison Method for Proving Inequalities": $3 \sum a^{3} b \leqslant\left(\sum a^{2}\right)^{2}$, are of equal strength; inequality (2) and $a^{2} b+b^{2} c+c^{2} a+a b c \leqslant \frac{4}{27}(a+b+c)^{3}$ are also of equal strength.

Additionally, inequalities (1) and (2) can also be proven using the incremental comparison method.