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
Your solution is right, but if you want to get all points for this problem, you need to explain, why it's enough to prove the starting inequality for equality case of two variables. It's not enough to say that it's true by uvw.
After words: $f$ decreases you can write the following.
Thus, it's enough to prove our inequality for a maximal value of $w^3$.
Now, $a$, $b$ and $c$ are roots of the equation $$(x-a)(x-b)(x-c)=0$$ or $$x^3-3ux^2+3v^2x-w^3=0$$ or $$g(x)=w^3,$$ where $$g(x)=x^3-3ux^2+3v^2x.$$ We see that $$g'(x)=3x^2-6ux+3v^2=3\left(x-(u-\sqrt{u^2-v^2})\right)\left(x-(u+\sqrt{u^2-v^2})\right),$$ which says that $g$ gets a local maximum for $x=u-\sqrt{u^2-v^2}$ and the equation $$g(x)=w^3$$ has three positive roots for $$\max\left\{0,g\left(u+\sqrt{u^2+v^2}\right)\right\}<w^3\leq g\left(u-\sqrt{u^2-v^2}\right)$$ and $w^3$ gets a maximal value, when the graph of $y=w^3$ is a tangent line to the graph of $g$, which happens for equality cases of two variables.
We got even that these variables should be equal to $u-\sqrt{u^2-v^2}.$
The rest you wrote already.