The Buffalo Way works.
After homogenization, it suffices to prove that $f(a,b,c)\ge 0$ where $f(a,b,c)$ is a polynomial given by
\begin{align}
f(a,b,c) &= 64abc(7a+b)^2(7b+c)^2(7c+a)^2\\
&\quad \times\left(\frac{3}{64}\frac{a+b+c}{abc} - \left(\frac{1}{7a+b} + \frac{1}{7b+c} + \frac{1}{7c+a}\right)^2\right).
\end{align}
WLOG, assume that $c = \min(a, b, c).$ There are two possible cases:
1) $c \le b\le a$: Let $c = 1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$.
$f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. So $f(1+s+t, 1+s, 1)\ge 0.$
2) $c \le a\le b$: Let $c =1, \ a=1+s, \ b=1+s+t; \ s,t\ge 0$. We have
\begin{align}
f(1+s, 1+s+t, 1) = a_5t^5 + a_4t^4 + a_3t^3 + a_2t^2 + a_1t + a_0
\end{align}
where
\begin{align}
a_5 &= 147\, s^2 - 784\, s + 6272,\\
a_4 &= 2940\, s^3 - 16583\, s^2 + 53648\, s + 82432 ,\\
a_3 &= 19551\, s^4 - 94494\, s^3 - 65760\, s^2 + 185344\, s + 139264,\\
a_2 &= 49686\, s^5 - 68407\, s^4 - 242656\, s^3 + 13824\, s^2 + 220160\, s + 81920,\\
a_1 &= 51744\, s^6 + 97584\, s^5 + 88848\, s^4 + 173056\, s^3 + 211968\, s^2 + 81920\, s ,\\
a_0 &= 81920\, s^2 + 270336\, s^3 + 344576\, s^4 + 224640\, s^5 + 87296\, s^6 + 18816\, s^7.
\end{align}
It is easy to obtain that $a_5, a_4, a_1, a_0 \ge 0$. Thus, we have
$$f(1+s, 1+s+t, 1)\ge (2\sqrt{a_5a_1} + a_3)t^3 + (2\sqrt{a_4a_0} + a_2)t^2.$$
It suffices to prove that $2\sqrt{a_5a_1} + a_3 \ge 0$ and $2\sqrt{a_4a_0} + a_2 \ge 0$.
Note that
\begin{align}
2\sqrt{a_5a_1} + a_3 &= \Big(2\sqrt{a_5a_1} - \frac{1}{3}\cdot 94494\, s^3 - \frac{1}{3} \cdot 65760\, s^2\Big)\\
&\quad + \Big(a_3 + \frac{1}{3}\cdot 94494\, s^3 + \frac{1}{3} \cdot 65760\, s^2\Big)
\end{align}
and
\begin{align}
2\sqrt{a_4a_0} + a_2 &= \Big(2\sqrt{a_4a_0} - \frac{1}{2}\cdot 68407\, s^4 - \frac{1}{2}242656\, s^3\Big)\\
&\quad + \Big(a_2 + \frac{1}{2}\cdot 68407\, s^4 + \frac{1}{2}242656\, s^3\Big).
\end{align}
It suffices to prove that
\begin{align}
2\sqrt{a_5a_1} - \frac{1}{3}\cdot 94494\, s^3 - \frac{1}{3} \cdot 65760\, s^2 &\ge 0,\\
a_3 + \frac{1}{3}\cdot 94494\, s^3 + \frac{1}{3} \cdot 65760\, s^2&\ge 0,\\
2\sqrt{a_4a_0} - \frac{1}{2}\cdot 68407\, s^4 - \frac{1}{2}242656\, s^3 &\ge 0,\\
a_2 + \frac{1}{2}\cdot 68407\, s^4 + \frac{1}{2}242656\, s^3 &\ge 0.
\end{align}
All of them can be reduced to polynomial inequalities in $s$ and not hard to prove. This completes the proof.
By AM-GM $$\sum_{cyc}\sqrt{\frac{a+b}{b+1}}\geq3\sqrt[6]{\prod\limits_{cyc}\frac{a+b}{a+1}}.$$
Thus, it's enough to prove that
$$(a+b)(a+c)(b+c)(3abc+1)^2\geq16a^2b^2c^2(a+1)(b+1)(c+1).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$(9uv^2-w^3)(3w^3+1)^2\geq16w^6(w^3+3v^2+3u+1)$$ and since by AM-GM $uv^2\geq w^3,$ it's enough to prove that
$$uv^2(3w^3+1)^2\geq2w^6(w^3+3v^2+3u+1),$$ which is true by AM-GM.
Indeed, by AM-GM $$2w^6(w^3+3v^2+3u+1)=2w^9+6v^2w^6+6uw^6+2w^6\leq$$
$$\leq2uv^2w^6+6uv^2w^5+6uv^2w^4+2uv^2w^3$$ and it's enough to prove that:
$$(3w^3+1)^2\geq2w^6+6w^5+6w^4+2w^3$$ or
$$(3w^3+1)^2\geq2w^3(w+1)^3.$$
Can you end it now?
Best Answer
Define $$ f(a,\lambda) = -\frac{a}{a^{11}+1} + \lambda \log(a) + \frac{1}{2} $$ Then, for any choice of $\lambda$, $$ f(a,\lambda) + f(b,\lambda) + f(c,\lambda) = -\frac{a}{a^{11}+1} -\frac{b}{b^{11}+1} -\frac{c}{c^{11}+1} + \frac{3}{2} $$ and we need to show that this is $\ge 0$.
It suffices to show that, for some $\lambda^*$ and for all $a$, $f(a, \lambda^*) \ge 0$.
Clearly, for any lambda, $f(a=1,\lambda) = 0$. In order to keep $f(a,\lambda) $ positive for $a >1$ and $a <1$, we demand
$$ 0 = \frac{d f(a,\lambda)}{d a}|_{a=1} $$
which results in $\lambda^* = - \frac94$. We therefore investigate
$$ f(a,\lambda^*) = -\frac{a}{a^{11}+1} -\frac{9}{4} \log(a) + \frac{1}{2} $$ By inspection, we have that $f(a,\lambda^*) \ge 0$ for $a\in (0, 1.1]$. So the inequality is obeyed at least for $a,b,c < 1.1$, and it remains to be shown that the inequality is obeyed outside this specification.
This gives rise to three cases:
case 1: $a,b,c > 1.1$. This is not possible since $abc = 1$.
case 2: $a < 1.1$ ; $b,c > 1.1$. Now observe two facts:
By inspection, $ \frac{a}{a^{11}+1} < 0.75$ for any $a$.
For $b > 1.1$, $ \frac{b}{b^{11}+1} \le \frac{1.1}{1.1^{11}+1} \simeq 0.2855$ since $ \frac{b}{b^{11}+1}$ is falling for $b > 1.1$.
Hence, in case 2, $ \frac{a}{a^{11}+1} + \frac{b}{b^{11}+1}+ \frac{c}{c^{11}+1} < 0.75 + 2\cdot 0.2855 = 1.3210 < \frac32$ which proves case 2.
case 3: $a,b < 1.1$ ; $c > 1.1$. Here $abc = 1$ requires $a\cdot b =1/c < 1.1^{-1} = 0.909$. Also note that, for some given $c$, $1/(1.1 c) <a<1.1$ in order to observe $a,b < 1.1$. Following case 2, we have that $f(c) = \frac{c}{c^{11}+1} $ is falling with $c$. These conditions could be further exploited (this has not yet been pursued in the comments).
As Martin R. poined out, the maximum will be attained at a point where at least two out of $a,b,c$ equal. In this case, this would be $a=b$. So we can consider proving $$ g(a) = \frac32 - \frac{2 a}{a^{11}+1} - \frac{a^{-2}}{a^{-22}+1} \ge 0 $$ for $a < 1/\sqrt{1.1} \simeq 0.9535$.
Note that in this range, the minimum of $g(a)$ occurs at $a^*\simeq 0.8385$ and has a value of $g(a^*) \simeq 0.00525$. Other than this inspection of the function $g(a)$, I couldn't offer a better proof.