Proof. By the AM-GM inequality,
$$a+b+\sqrt{2 a+2 c}=(a+b)+\sqrt{\frac{a+c}{2}}+\sqrt{\frac{a+c}{2}} \geqslant 3 \sqrt[3]{\frac{(a+b)(a+c)}{2}}$$

Thus,
$$\left(\frac{1}{a+b+\sqrt{2 a+2 c}}\right)^{3} \leqslant \frac{2}{27(a+b)(a+c)}$$

Similarly, we have
$$\left(\frac{1}{b+c+\sqrt{2 b+2 a}}\right)^{3} \leqslant \frac{2}{27(b+c)(b+a)},\left(\frac{1}{c+a+\sqrt{2 c+2 b}}\right)^{3} \leqslant \frac{2}{27(c+a)(c+b)}$$

Therefore, it suffices to prove that
$$\sum_{c y c} \frac{2}{27(a+b)(a+c)} \leqslant \frac{8}{9}$$

This is equivalent to
$$6(a+b)(b+c)(c+a) \geqslant a+b+c$$

That is,
$$6 a b c\left(\sum a\right)\left(\sum a b\right) \geqslant a b c \sum a+6 a^{2} b^{2} c^{2}$$

Notice that $16(a+b+c) \geqslant \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, i.e., $16 a b c \sum a \geqslant \sum a b$, so
$$16 a b c \sum a\left(\sum a b\right) \geqslant\left(\sum a b\right)^{2} \geqslant 3 a b c \sum a$$

Additionally, by the AM-GM inequality, we have
$$2 a b c \sum a \sum a b \geqslant 18 a^{2} b^{2} c^{2}$$

Therefore, adding (4) and (5) and dividing by 3 yields (3).
Thus, the inequality is proved, with equality holding if and only if $a=b=c=\frac{1}{4}$.