<s>[INST] 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.
 [/INST] 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. 
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