Here's a proof.
Following the task description, we want to establish
$$
\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}
$$
We will follow two paths for separate cases.
Path 1:
Clearing denominators, we obtain
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z)+ (z-x)^2 (z+x) \geq \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2} (x+y) (y+z) (z+x)
$$
Now we use the general formula (by AM-GM):
$$
(a+b+c)^3 \geq 27 a b c
$$
Applying this to $a = (x+y); b = (y+z); c = (z+x)$ gives
$$
(x+y+z)^3 \geq \frac{27}{8} (x+y) (y+z) (z+x)
$$
It then suffices to show
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq \frac{(y-z)^2}{(x+y+z)^2} (x+y+z)^3
$$
which is
$$
(x-y)^2 (x+y) + (z-x)^2 (z+x) \geq (y-z)^2 x
$$
Since
$$
(y-z)^2 = (y-x + x- z)^2 = (y-x)^2 + (x- z)^2 + 2 (y-x)(x-z)
$$
this translates into
$$
(y-x)^2 y + (x-z)^2 z \geq 2 x(y-x)(x-z)
$$
For the two cases $y\geq x ; z\geq x $ and $y\leq x ; z\leq x $ the RHS $\leq 0$ so we are done. For the other two cases, by symmetry, it remains to show the case $y> x ; z < x $.
Rearranging terms, we can also write
$$
(y-x)^3 - (x-z)^3 +x ((y - x) + (z-x))^2 \geq 0
$$
This holds true at least for $(y-x)^3 \geq (x-z)^3$ or $y+z\geq 2 x$.
So the proof is complete other than for the case $y+z < 2x$ and [ $y> x ; z < x $ or $z> x ; y < x $ ].
Path 2.
For the remaining case $y+z < 2 x $ and [$y> x ; z < x$ or $z> x ; y < x $] we will follow a different path. Again, by symmetry, we must inspect only $y+z < 2x$ and $y> x > z$.
Clearing all denominators gives
$$
8 (x+y+z)^2 \sum_{cyc} {(x-y)^2}{(x+y)} - 27 (y-z)^2 (x+y)(y+z) (z+x) \geqslant 0
$$
By homogeneity, we set $y=1+z$.
The condition $y+z < 2x$ then translates into $1+2z < 2x$, hence we further set $x = z + (1 +q)/2$ where $0\leq q \leq 1$ since also $x = z + (1 +q)/2 < y = 1 +z$.
This gives a lengthy expression
$$
8 (3 z + 3/2 + q/2)^2 \left\{ ((q-1)/2)^2 (2z + (3 +q)/2) + ((1 +q)/2 )^2(2z +(1 +q)/2 ) + (1 + 2z) \right\} - 27 (2z + (3 +q)/2) (2z +(1 +q)/2 )(1 + 2z)\geqslant 0
$$
The LHS is a third order expression in $z$ with leading (in $z^3$ ) term $72 q^2 z^3$, hence for large enough $z$ it is rising with $z$. A remarkable feature of this expression is that for the considered range $0\leq q \leq 1$ it is actually monotonously rising for all $z$. To see this, consider whether there are points with zero slope. The first derivative of the expression with respect to $z$ is
$$
216 q^2 z^2 + (84 q^3 + 216 q^2 - 108 q) z + (81 q^2)/2 + 42 q^3 + 8 q^4 - 54 q + 27/2
$$
Equating this to zero gives
$$
z_{1,2} = -(18 q \pm q \sqrt{q^2 - 45} + 7 q^2 - 9)/(36 q)
$$
however there are no real roots in the considered range $0\leq q \leq 1$, so monotonicity (rising with $z$) is established.
Hence, to show the inequality it suffices to inspect the LHS at the smallest $z=0$. This gives
$$
q^5/2 + 4 q^4 + 10 q^3 + (9 q^2)/4 - (27 q)/2 + 27/4 \geq 0
$$
An even stronger requirement is
$$
f(q) =
10 q^3 + (9 q^2)/4 - (27 q)/2 + 27/4 \geq 0
$$
In the considered range $0\leq q \leq 1$, $f(q)$ has a minimum which is obtained by taking the first derivative,
$$
30 q^2 + (9 q)/2 - 27/2
$$
and equating to zero, which gives $q = 3/5$, and the above $f(q)$ then gives
$$
f(q = 3/5) = 81/50
$$
This establishes the inequality. $ \qquad \Box$
