Let $S$ be the sum

$$\frac{5+2 b+c^{2}}{1+a}+\frac{5+2 c+a^{2}}{1+b}+\frac{5+2 a+b^{2}}{1+c}$$

Let $x_{i}$ be the $i^{\text {th }}$ smallest element of the set $\{a, b, c\}$. The rearrangement inequality indicates that

$$\begin{aligned} & \frac{b}{1+a}+\frac{c}{1+b}+\frac{a}{1+c} \geqslant \frac{x_{1}}{1+x_{1}}+\frac{x_{2}}{1+x_{2}}+\frac{x_{3}}{1+x_{3}} \text { and } \\ & \frac{c^{2}}{1+a}+\frac{a^{2}}{1+b}+\frac{b^{2}}{1+c} \geqslant \frac{x_{1}^{2}}{1+x_{1}}+\frac{x_{2}^{2}}{1+x_{2}}+\frac{x_{3}^{2}}{1+x_{3}}, \end{aligned}$$

so that

$$\begin{aligned} S & \geqslant \frac{5+2 x_{1}+x_{1}^{2}}{1+x_{1}}+\frac{5+2 x_{2}+x_{2}^{2}}{1+x_{2}}+\frac{5+2 x_{3}+x_{3}^{2}}{1+x_{3}} \\ & \geqslant 4\left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\frac{1}{1+x_{3}}\right)+\left(1+x_{1}\right)+\left(1+x_{2}\right)+\left(1+x_{3}\right) \\ & \geqslant 4+4\left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\frac{1}{1+x_{3}}\right) \end{aligned}$$

The function $\mathrm{f}: \mathrm{x} \mapsto 1 /(1+x)$ being convex, we also know that

$$\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\frac{1}{1+x_{3}}=f\left(x_{1}\right)+\mathbf{f}\left(x_{2}\right)+\mathbf{f}\left(x_{3}\right) \geqslant 3 \mathbf{f}\left(\frac{x_{1}+x_{2}+x_{3}}{3}\right)=3 \mathbf{f}(1 / 3)=9 / 4$$

We deduce that $S \geqslant 4+9=13$.