Asymmetric inequality in three variables $\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$ Consider three real positive variables $a,\ b$ and $c$. Prove that the following inequality holds:
$$\frac{3(a+b)^2(b+c)^2}{4ab^2c} \geq 7+\frac{5(a^2+2b^2+c^2)}{(a+b)(b+c)}$$
My progress: We can prove that both sides are greater than $12$ using AM-GM:
$$LHS \geq \frac{3 \cdot 4ab \cdot 4bc}{4a^2bc} = 12$$
and 
$$RHS \geq 7+\frac{5[(a+b)^2+(b+c)^2]}{2(a+b)(b+c)} \geq 7+5 = 12$$
So, substract $12$ from both sides and write the inequality into:
$$3\cdot \frac{(a+b)^2(b+c)^2-16ab^2c}{4ab^2c} \geq 5 \cdot \frac{a^2+b^2+c^2-ab-bc-ca}{(a+b)(b+c)}$$
or
$$3\cdot \frac{(b+c)^2(a-b)^2+4ab(b-c)^2}{4ab^2c} \geq \frac{5}{2}\cdot \frac{(a-b)^2+(b-c)^2+(c-a)^2}{(a+b)(b+c)}$$
My next idea was to use $(c-a)^2\leq 2[(a-b)^2+(b-c)^2]$ and write it into a sum of square form with only $(a-b)^2$ and $(b-c)^2$. However, I couldn't reach significant progress.

Making the coordinates transformation
$$
\cases{
\frac{a+b}{a}=x\\
\frac{b+c}{c}=y\\
a b^2 c=z
}
$$
solving for $a,b,c$ we have
$$
\left\{
\begin{array}{rcl}
 a & = & \frac{z}{(y-1)^2 \left(\frac{(x-1) z}{(y-1)^3}\right)^{3/4}} \\
 b & = & (y-1) \sqrt[4]{\frac{(x-1) z}{(y-1)^3}} \\
 c & = & \sqrt[4]{\frac{(x-1) z}{(y-1)^3}} \\
\end{array}
\right.
$$
conditioned to $x > 1,\ y > 1,\ z>0$. Substituting into
$$
3\frac{(a+b)^2(b+c)^2}{4ab^2c}- 7 - 5\frac{a^2+2b^2+c^2}{(a+b)(b+c)}\ge 0
$$
we have
$$
f(x,y) = \frac{3 x^3 y^3-4 x^2 (y (17 y-27)+15)+4 x (y (27 y-47)+30)-20 (3 (y-2) y+4)}{4 (x-1) x (y-1) y}\ge 0
$$
Now $f(x,y)$ has a minimum for $x=y=2$ such that $f(2,2) = 0$
Follows a plot of $f(x,y)$

