Without loss of generality, assume that $a \geq b \geq c$. From the condition, we get $a \geq 1 \geq c > 0$. Write the inequality as
$$\frac{a+b+c}{c}+\frac{a+b+c}{b}+\frac{a+b+c}{c} \geq a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3$$
which is equivalent to
$$(a+b+c-1)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right) \geq 4$$
Since $(a-1)(1-c) \geq 0$, we obtain $a+c \geq 1+a c$ and thus $\frac{1}{a}+\frac{1}{c} \geq 1+\frac{1}{a c}$. Therefore,
$$\begin{aligned} & (a+b+c-1)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right) \\ & \geq (b+a c)\left(\frac{1}{b}+\frac{1}{a c}\right)=2+\frac{b}{a c}+\frac{a c}{b} \geq 2+2=4 \end{aligned}$$
This completes the proof. Equality holds for $a=b=c=1$.