To prove this problem, we notice the following fact: replacing $a, b, c$ with $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$ does not change the inequality, so we can assume $0<a, b, c<1, a+b+c=1$. Then, $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}=\frac{(a+1)^{2}}{2 a^{2}+(1-a)^{2}}=\frac{(a+1)^{2}}{3 a^{2}-2 a+1}$. Let $f(x)=\frac{(x+1)^{2}}{3 x^{2}-2 x+1}, 0<x<1$, the tangent line at $x=\frac{1}{3}$ is $g(x)=\frac{12 x+4}{3}$. Therefore, $f(x)-g(x)=$ $\frac{-36 x^{3}+15 x^{2}+2 x-1}{3\left(3 x^{2}-2 x+1\right)}=\frac{-(3 x-1)^{2}(4 x+1)}{3\left(3 x^{2}-2 x+1\right)} \leqslant$ 0. So, $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant \frac{12 a+4}{3}$. Similarly, $\frac{(a+2 b+c)^{2}}{2 b^{2}+(a+c)^{2}} \leqslant \frac{12 b+4}{3}, \frac{(a+b+2 c)^{2}}{2 c^{2}+(b+a)^{2}}$ $\frac{12 c+4}{3}$. Adding these inequalities yields the desired result.