Problem 1302

Proof: Let $f(x)=\frac{x^{2}+9}{3 x^{2}-6 x+9}, x \in(0,3)$, then $f^{\prime}(x)=\frac{-6\left(x^{2}+6 x-9\right)}{\left(3 x^{2}-6 x+9\right)^{2}}=$ $\frac{-2\left(x^{2}+6 x-9\right)}{3\left(x^{2}-2 x+3\right)^{2}}$. Since $f^{\prime}(x)$ is not monotonic, we cannot handle it using concavity or convexity. Noting the geometric meaning of the derivative:
$f^{\prime}\left(x_{0}\right)=\frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}$, we have $f(x)=f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)+f\left(x_{0}\right)$, we can consider whether there exists a bounding function for $f(x)$ at $x=x_{0}$. Since the function's extremum is obtained when the variable $x$ takes the average value, we can consider the tangent function of $f(x)$ at $x=1$, which is $y=f^{\prime}(1)(x-1)+f(1)=\frac{x+4}{3}$.
Because $\frac{x^{2}+9}{3 x^{2}-6 x+9}-\frac{x+4}{3}=\frac{\left(x^{2}+9\right)-(x+4)\left(x^{2}-2 x+3\right)}{3 x^{2}-6 x+9}$
$$=\frac{-(x+3)(x-1)^{2}}{3 x^{2}-6 x+9} \text {, }$$

Therefore, when $0<x<3$, $\frac{x^{2}+9}{3 x^{2}-6 x+9} \leqslant \frac{x+4}{3}$.
Thus, $\frac{a^{2}+9}{2 a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2 b^{2}+(c+a)^{2}}+\frac{c^{2}+9}{2 c^{2}+(a+b)^{2}}$
$$\begin{array}{l} =\frac{a^{2}+9}{3 a^{2}-6 a+9}+\frac{b^{2}+9}{3 b^{2}-6 b+9}+\frac{c^{2}+9}{3 c^{2}-6 c+9} \\ \leqslant \frac{a+4}{3}+\frac{b+4}{3}+\frac{c+4}{3}=5 . \end{array}$$