Solution. The inequality can be rewritten as
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \leq 7$$

WLOG, we may assume that $a \geq b \geq c$, then
$$(a-b)(a-c) \geq 0 \Rightarrow\left\{\begin{array}{l} \frac{a}{c}+1 \geq \frac{a}{b}+\frac{b}{c} \\ \frac{c}{a}+1 \geq \frac{c}{b}+\frac{b}{a} \end{array}\right.$$
which implies that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{b}+\frac{b}{a} \leq \frac{a}{c}+\frac{c}{a}+2$$

We conclude
$$\Rightarrow \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \leq 2+2\left(\frac{a}{c}+\frac{c}{a}\right)=7-\frac{(a-2 c)(2 a-c)}{a c} \leq 7$$
because $2 c \geq a \geq c$. Equality holds for $(a, b, c)=(2,2,1)$ or $(2,1,1)$ or permutations.