SOLUTION. Similarly to the previous proofs, we assume first that $a \geq b \geq c$ and denote
$$f(a, b, c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{8}{a+b+c}-\frac{6}{\sqrt{a b+b c+c a}}$$

Let $t=b+c$, then we have
$$\begin{aligned} f(a, b, c)-f(a, t, 0) & =\frac{1}{a+b}+\frac{1}{a+c}-\frac{1}{a+b+c}-\frac{1}{a}-\frac{6}{\sqrt{a b+b c+c a}}+\frac{6}{\sqrt{a(b+c)}} \\ & \geq-\frac{c}{a(a+c)}+\frac{6(\sqrt{a b+b c+c a}-\sqrt{a b+a c})}{\sqrt{a(b+c)(a b+b c+c a)}} \end{aligned}$$

Using $a b+b c+c a \leq(a+c)^{2}$ and $a(b+c) \leq 2 a b$, we get
$$\begin{array}{c} \frac{\sqrt{a b+b c+c a}-\sqrt{a b+a c}}{\sqrt{a(b+c)(a b+b c+c a)}}=\frac{b c}{\sqrt{a(b+c)}(a b+b c+c a)+a(b+c) \sqrt{a b+b c+c a}} \\ \quad \geq \frac{b c}{3 a b \sqrt{a(b+c)}+2 a b(a+c)}=\frac{c}{a(3 \sqrt{a(b+c)}+2(a+c))} \end{array}$$

Therefore
$$f(a, b, c)-f(a, t, 0) \geq \frac{6 c}{a(3 \sqrt{a(b+c)}+2(a+c))}-\frac{c}{a(a+c)}$$

Moreover, since $a \geq b \geq c$, we infer that
$$6(a+c)=4(a+c)+2(a+c) \geq \frac{3}{2}(a+b+c)+2(a+c) \geq 3 \sqrt{a(b+c)}+2(a+c)$$
which means that $f(a, b, c) \geq f(a, t, 0)$. Furthermore, by AM-GM inequality, we conclude
$$\begin{array}{l} f(a, t, 0)=\frac{9}{a+t}+\frac{1}{a}+\frac{1}{t}-\frac{6}{\sqrt{a t}} \\ =\frac{1}{\sqrt{a t}}\left(\frac{9 \sqrt{a t}}{a+t}+\frac{a+t}{\sqrt{a t}}-6\right) \geq 0 \end{array}$$

This ends the proof. The equality holds for $a+t=3 \sqrt{a t}$ or $(a, b, c) \sim$ $\left(\frac{7 \pm \sqrt{45}}{2}, 1,0\right)$