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.
Best Answer
Titu's lemma gives us
Since $(p+q+r) ^2 \geq 3 (pq+qr+rs)$, so $3 \geq pq+qr+rs $ and thus
$$ \sum \frac{q}{ 3 + p } \geq \frac{9}{9 + pq+qr+rs} \geq \frac{3}{4}.$$
The original problem could be approach in as similar manner.
Then, since $ \frac{1}{3} \geq ab+bc+ca$, thus
$$\sum \frac{7+2b}{1+a} \geq \frac{23^2}{30 + 2ab+2bc+2ca} \geq \frac{69}{4}.$$