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.
 A: Partial solution. I hope, it can help.
Let $a+c=2p$ and $ac=q^2,$ where $q>0.$
Thus, by AM-GM $p\geq q$ and we need to prove that
$$\frac{3(b^2+2bp+q^2)^2}{4q^2b^2}\geq7+\frac{5(b^2+4p^2-2q^2)}{b^2+2pb+q^2}.$$
Now, consider two cases:


*

*$b\geq q$, $p=q+u$ and $b=q+v$.


Thus, we need to prove that:
$$72uq^5+4(16u^2+94uv+19v^2)q^4+8(3u^3+34u^2v+85uv^2+19v^3)q^3+$$
$$+4v(18u^3+97u^2v+130uv^2+28v^3)q^2+18v^2(2u+v)^2(u+2v)q+3v^3(2u+v)^3\geq0,$$
which is obviously true.


*$b\leq q$, $q=b+u$, $p=b+u+v$.


Thus, we need to prove that:
$$72vb^5+4(19u^2-4uv+16v^2)b^4+8(19u^3-13u^2v-2uv^2+3v^3)b^3+$$
$$+4u^2(28u^2-4uv-11v^2)b^2+18u^4(2u+v)b+3u^6\geq0.$$ 
Now, we see that it's a cubic inequality of $v$ and after using a derivative we can get a minimal point and to end a proof. 
A: 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 $f(x,y)$

A: Another proof:
Since the inequality is symmetric in $a$ and $c$, WLOG, assume that $a \ge c$.
Due to homogeneity, assume that $c = 1$.
Let $a = 1 + s$ for $s \ge 0$.
We split into two cases:
1) $0 < b \le 1$: Let $b = \frac{1}{1+t}$ for $t \ge 0$. We have
\begin{align}
\mathrm{LHS} - \mathrm{RHS} &= \frac{1}{4(2+t)(st+s+t+2)(1+s)(1+t)^2}f(s,t) 
\end{align}
where
\begin{align}
f(s,t) &= 3 s^3 t^6+27 s^3 t^5+9 s^2 t^6+79 s^3 t^4+90 s^2 t^5+9 s t^6+109 s^3 t^3\\
&\quad +281 s^2 t^4+99 s t^5+3 t^6+78 s^3 t^2+412 s^2 t^3+314 s t^4+36 t^5\\
&\quad +28 s^3 t+324 s^2 t^2+452 s t^3+112 t^4+4 s^3+140 s^2 t+304 s t^2\\
&\quad +152 t^3+28 s^2+76 s t+76 t^2.
\end{align}
Clearly, $f(s,t) \ge 0$. True.
2) $b > 1$: Let $b = 1+r$ for $r > 0$. We have
\begin{align}
\mathrm{LHS} - \mathrm{RHS} &= \frac{1}{4(1+s)(1+r)^2(2+s+r)(2+r)}g(s,r)
\end{align}
where
\begin{align}
g(s, r) &= 3 r^6+9 r^5 s+9 r^4 s^2+36 r^5+22 r^4 s+44 r^3 s^2\\
&\quad +112 r^4+4 r^3 s+44 r^2 s^2+76 r^3+9 r s^2+9 s^2+\tfrac{15}{16} s^3\\
&\quad + 19 (2 r-s)^2+19 r (2 r-s)^2+3 (r-1)^2 r s^3+4 s^3 (r-\tfrac78)^2.
\end{align}
Clearly, $g(s,r) > 0$. True.
We are done.
