Proving $\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $ For $a,b,c \in (0,\infty).$ Prove$:$
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $$
My proof by SOS$:$
$$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{ 2} \right) \left( a+b \right) \left( b+c \right) \left( c+a \right)\, \cdot \,(\text{LHS}-\text{RHS})$$
$$=\frac{5}{4} abc \sum\limits_{cyc} c^2 (a+b-2c)^2 (a-b)^2 +\frac{1}{4} \sum\limits_{cyc} {c}^{3} \left( 4\,{a}^{2}+3\,ab+4\,{b}^{2} \right) \left( a-b
\right) ^{4}$$
However$,$ it's hard to find this SOS's form without computer. 
So I am looking for alternative solution without $uvw.$ Thanks very much!
 A: We have
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab} - 1 - \frac{16abc}{(a+b)(b+c)(a+c)}$$
$$=\sum{\frac{(c-a)^2(c-b)^2}{(c^2+ab)(b+c)(c+a)}}\geqslant 0.$$
A: We can prove this inequality as an inequality in the Schur's form typing. 
Indeed, we need to prove that:
$$\sum_{cyc}\frac{a(b+c)}{a^2+bc}\geq1+\frac{16abc}{\prod\limits_{cyc}(a+b)}$$ or
$$2-\frac{16abc}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\left(1-\frac{a(b+c)}{a^2+bc}\right)$$ or
$$\frac{2\sum\limits_{cyc}(a^2b+a^2c-2abc)}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}$$ or
$$\frac{2\sum\limits_{cyc}(b+c)(a-b)(a-c)}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}$$ or
$$\sum_{cyc}(a-b)(a-c)\left(\frac{2}{(a+b)(a+c)}-\frac{1}{a^2+bc}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(a-c)^2}{(a+b)(a+c)(a^2+bc)}\geq0$$ and we are done! 
A: $$LHS-RHS=\frac{4abc(b-c)^{2}(c-a)^{2}(a-b)^{2}}{(b+c)(c+a)(a+b)(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}+\frac{(a^{2}+b^{2}+c^{2}-bc-ca-ab)(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}-a^{2}bc-b^{2}ca-c^{2}ab)}{(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}\geq 0.$$
