Prove $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32+ \frac{27}{16}\frac{(y-z)^2}{(x+y+z)^2}$ 
$x,y,z >0$, prove
  $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32+ \frac{27}{16}\frac{(y-z)^2}{(x+y+z)^2}$$

This inequality is easier than the other one. Previously, I learned from Jack D'Aurizio at this post, so my first step is to use the Lagrange's identity. I get
$$\sum_{cyc}\frac{x}{y+z}=\frac32+\frac12\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)}$$ 
Thus it remains to prove
$$\tag{1}\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}$$
I spent several hours to prove (1) but not success. Please help.

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$
