69. Since
$$\begin{array}{l} 2\left(a^{7}+b^{7}\right)-\left(a^{5}+b^{5}\right)\left(a^{2}+b^{2}\right)=\left(a^{5}-b^{5}\right)\left(a^{2}-b^{2}\right)= \\ (a-b)^{2}(a+b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right) \geqslant 0 \end{array}$$

Therefore, $\frac{a^{7}+b^{7}}{a^{5}+b^{5}} \geqslant \frac{a^{2}+b^{2}}{2}$, similarly, $\frac{b^{7}+c^{7}}{b^{5}+c^{5}} \geqslant \frac{b^{2}+c^{2}}{2}, \frac{c^{7}+a^{7}}{c^{5}+a^{5}} \geqslant \frac{c^{2}+a^{2}}{2}$.
Adding these inequalities, we get $\frac{a^{7}+b^{7}}{a^{5}+b^{5}}+\frac{b^{7}+c^{7}}{b^{5}+c^{5}}+\frac{c^{7}+a^{7}}{c^{5}+a^{5}} \geqslant a^{2}+b^{2}+c^{2} \geqslant \frac{1}{3}(a+b+c)^{2}=\frac{1}{3}$.