74.
$$\begin{aligned} a^{2}+\left(\frac{a+b+c}{3}\right)^{2} & \geqslant \frac{2 a(a+b+c)}{3} \\ b^{2}+\left(\frac{a+b+c}{3}\right)^{2} & \geqslant \frac{2 b(a+b+c)}{3} \\ c^{2}+\left(\frac{a+b+c}{3}\right)^{2} & \geqslant \frac{2 c(a+b+c)}{3} \end{aligned}$$

Adding them up, we get $a^{2}+b^{2}+c^{2} \geqslant \frac{1}{3}(a+b+c)^{2}$, thus the left inequality is proved.
$$\begin{array}{l} \text { By }(x+y+z)^{2} \geqslant 3(x y+y z+z x) \text { we get } \\ \quad\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)^{2} \geqslant 3 a^{2} b^{2} c^{2}\left(a^{2}+b^{2}+c^{2}\right) \end{array}$$

That is $\square$
$$a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geqslant a b c \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}$$

Dividing both sides by $3 a b c$ we get
$$\frac{1}{3}\left(\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}\right) \geqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}$$

In summary
$$\frac{1}{3}(a+b+c) \leqslant \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \leqslant \frac{1}{3}\left(\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b}\right)$$