Solution 1. Adding $3$ to each side, we have to show
$$\begin{aligned} 5 &\le \frac{a+b}{c+2} + 1 + \frac{b+c}{a+2} + 1 + \frac{c+a}{b+2} + 1 \\ &= (a+b+c+2) \left( \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \right) \\ &= 5 \left( \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \right) \end{aligned}$$
Dividing by $5$, this is equivalent to
$$\frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \ge 1 \quad (4)$$
Alternatively, we may substitute $a + b = 3 - c = 5 - (c + 2)$ in the first numerator and so on, which again leads us to (4). Now Jensen's inequality applied to the convex function $f(x) = \frac{1}{x+2}$ at the points $a$, $b$, $c$ implies:
$$\frac{1}{3} \left( \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \right) \ge \frac{1}{\frac{1}{3}(a+b+c)+2} = \frac{1}{3}.$$
Equality holds (in general) when $a = b = c$ and in this specific case when $a = b = c = 1$ since we are given $a + b + c = 3$.

Rather than invoking Jensen's inequality, (4) can be proved by appeal to the AM-HM inequality, the AM-GM inequality or the Cauchy-Schwarz inequality. The AM-HM inequality application is immediate after multiplication by $(a+2) + (b+2) + (c+2) = 9$. The use of AM-GM requires two steps:
$$\begin{aligned} \frac{1}{3} \left( \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \right) &\ge \frac{1}{\sqrt[3]{(a+2)(b+2)(c+2)}} \\ &\ge \frac{3}{a+2+b+2+c+2} = \frac{1}{3}. \end{aligned}$$
The Cauchy-Schwarz approach applies the inequality to the vectors
$$\left(\sqrt{a+2}, \sqrt{b+2}, \sqrt{c+2}\right) \quad \text{and} \quad \left(\frac{1}{\sqrt{a+2}}, \frac{1}{\sqrt{b+2}}, \frac{1}{\sqrt{c+2}}\right)$$
whence
$$\begin{aligned} 9 &\le ((a+2) + (b+2) + (c+2)) \left( \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \right) \\ &= 9 \left( \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \right). \end{aligned}$$

$$\frac{a+b}{5-a-b} + \frac{3-a}{a+2} + \frac{3-b}{b+2} \geq 2.$$
Cross-multiplying and moving all terms to the left (noting all the denominators are positive) gives: $5a^2b + 5ab^2 + 5a^2 + 5b^2 - 10ab - 15a - 15b + 20 \geq 0$. The left hand side can be rearranged as $5b(a-1)^2 + 5a(b-1)^2 + 5(a+b-2)^2$. This is non-negative, and zero iff $a = b = 1$.