To prove the desired inequality, we can write it as:
$$\begin{array}{l} \frac{1}{3}-\sum \frac{a^{2}}{(2 a+b)(2 a+c)}= \\ \sum\left(\frac{a}{3(a+b+c)}-\frac{a^{2}}{(2 a+b)(2 a+c)}\right)= \\ \sum \frac{1}{3(a+b+c)} \cdot \frac{a}{(2 a+b)(2 a+c)}(a-b)(a-c)= \\ \frac{1}{3(a+b+c)} \sum \frac{a}{(2 a+b)(2 a+c)}(a-b)(a-c) \end{array}$$

Therefore, we only need to prove:
$$x(a-b)(a-c)+y(b-c)(b-a)+z(c-a)(c-b) \geqslant 0$$

where $x=\frac{a}{(2 a+b)(2 a+c)}, y=\frac{b}{(2 b+c)(2 b+a)}, z=\frac{c}{(2 c+a)(2 c+b)}$. Without loss of generality, assume $a \geqslant b \geqslant c$. We prove $a x \geqslant b y \Leftrightarrow$
$$\begin{array}{l} a \frac{a}{(2 a+b)(2 a+c)} \geqslant b \frac{b}{(2 a+c)(2 b+a)} \Leftrightarrow \\ a(2 b+a) a(2 b+c) \geqslant b(2 a+b) b(2 a+c) \end{array}$$

And we also have:
$$a(2 b+a) \geqslant b(2 a+b), a(2 b+c) \geqslant b(2 a+c)$$

Therefore, $a x \geqslant b y$ holds, and by Theorem 5.1, the original inequality is proved.
The equality holds if and only if $a=b=c$.