Proving $\frac{1}{16} \sum \frac{(b+c)(c+a)}{ab} +\frac{9}{4} \geq 4\sum \frac{ab}{(b+c)(c+a)}$ For $a,b,c>0$. Prove: $$\frac{1}{16} \sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+\frac{9}{4} \geq 4\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}$$
My SOS's proof  is:
It's equivalent to: $$\frac{1}{27}\sum\limits_{cyc} ab \left( a+b-8\,c \right) ^{2} \left( a+b-2\,c \right) ^{2}+\frac{26}{27}\sum\limits_{cyc}ab \left( a-b \right) ^{2} \left( a+b-2\,c \right) ^{2} +{\frac{50}{27}} \Big[\sum\limits_{cyc} a(b-c)^2\Big]^2 \geq 0$$
However, it's hard to find without computer.
So I'm looking for alternative solution without $uvw$. Thanks for a real lot!
 A: We have$:$\begin{align*}\sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+36 - 64\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}
=\frac{\prod \,(a+b)\sum\limits_{cyc} a(a-b)(a-c)+\Big[\sum\limits_{cyc} c(a-b)^2\Big]^2}{a\,b\,c\,(a+b)\,(b+c)\,(c+a)}\geq 0\end{align*}
The last inequality is Schur degree $3,$ you can see many proof by SOS here.
Or this one is stronger$:$ here
A: We need to prove that:
$$\sum_{cyc}\left(\frac{(a+c)(b+c)}{ab}-4\right)\geq16\sum_{cyc}\left(\frac{4ab}{(a+c)(b+c)}-1\right)$$ or 
$$\sum_{cyc}(c^2+ac+bc-3ab)\left(\frac{1}{ab}+\frac{16}{(a+c)(b+c)}\right)\geq0$$ or
$$\sum_{cyc}((c-a)(3b+c)-(b-c)(2a+c))\left(\frac{1}{ab}+\frac{16}{(a+c)(b+c)}\right)\geq0$$ or
$$\sum_{cyc}(a-b)\left((3c+a)\left(\tfrac{1}{bc}+\tfrac{16}{(a+b)(a+c)}\right)-(3c+b)\left(\tfrac{1}{ac}+\tfrac{16}{(a+b)(b+c)}\right)\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(3(a+b)c^3+4(a^2-6ab+b^2)c^2+(a+b)(a^2+5ab+b^2)c+ab(a+b)^2)\geq0,$$ which is true because by AM-GM
$$3(a+b)c^3+4(a^2-6ab+b^2)c^2+(a+b)(a^2+5ab+b^2)c+ab(a+b)^2\geq$$
$$\geq6\sqrt{ab}c^3-16abc^2+14\sqrt{a^3b^3}c+4a^2b^2\geq0.$$
