Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$ For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$
NguyenHuyen gave the following expression$:$ 
$$\sum \frac12\, \left( 8\,{a}^{3}b+{a}^{3}c+8\,{a}^{2}{b}^{2}+11\,{a}^{2}bc+7\,a
{b}^{3}+13\,a{b}^{2}c+3\,ab{c}^{2}+3\,b{c}^{3}+2\,{c}^{4} \right) 
 \left( a+b \right) ^{2} \left( a-b \right) ^{2} \geqslant 0$$
My work with Titu's Lemma and Maple and by lucky!
By Titu's Lemma$,$ we have$:$
$$\text{LHS} \geqq \frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc} a^2(b+c)^2} +\frac{1}{4}\geqq \text{RHS}$$
Last inequality equivalent to$:$
$$\,{\frac {\sum\limits_{cyc} \left( a-b \right) ^{2}\Big[bc \left( 2\,{a}^{2}+ab+ca+{c}^{
2} \right) +2\,ac \left( a-c \right) ^{2}+2\,ab \left( {a}^{2}+{b}^{2}
\right) +c \left( b-c \right) ^{2} \left( a+2\,b \right)\Big]}{ 8\left( {
a}^{2}{b}^{2}+{a}^{2}bc+{a}^{2}{c}^{2}+a{b}^{2}c+ab{c}^{2}+{b}^{2}{c}^
{2} \right) \left( ab+ca+bc \right) }}\geqq 0$$
However$,$ it's very hard with me to find a nice SOS if not have Maple$,$ who have a simple proof for it?
Without $\it{uvw}$ and Buffalo Way if you can!
Thanks a lot!
$\lceil$You can also see here. $\rfloor$
 A: Yes, SOS helps!
Indeed, we need to prove that
$$\sum_{cyc}\left(\frac{a^2}{(b+c)^2}-\frac{1}{4}\right)\geq\frac{a^2+b^2+c^2}{ab+ac+bc}-1$$ or
$$\sum_{cyc}\frac{(2a+b+c)(a-b-(c-a))}{(b+c)^2}\geq\frac{2\sum\limits_{cyc}(a-b)^2}{ab+ac+bc}$$ or
$$\sum_{cyc}(a-b)\left(\frac{2a+b+c}{(b+c)^2}-\frac{2b+a+c}{(a+c)^2}\right)\geq\frac{2\sum\limits_{cyc}(a-b)^2}{ab+ac+bc}$$ or
$$\sum_{cyc}(a-b)^2\left(\frac{2a^2+2b^2+3c^2+3ab+5ac+5bc}{(a+c)^2(b+c)^2}-\frac{2}{ab+ac+bc}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+b)^2(2a^3b+a^2b^2+2ab^3+2a^3c+6a^2bc+6ab^2c+2b^3c+3a^2c^2+5abc^2+3b^2c^2-ac^3-bc^3-2c^4)\geq0,$$ for which it's enough to prove that:
$$\sum_{cyc}(a-b)^2(a+b)^2(2a^3c+2b^3c-2c^4+a^2c^2+2abc^2+b^2c^2-ac^3-bc^3)\geq0$$ or
$$2\sum_{cyc}(a-b)^2(a+b)^2c(a^3+b^3-c^3)+\sum_{cyc}(a-b)^2(a+b)^3c^2(a+b-c)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, $$b^2\sum_{cyc}(a-b)^2(a+b)^2c(a^3+b^3-c^3)\geq$$
$$b^2(a-c)^2(a+c)^2b(a^3+c^3-b^3)+b^2(b-c)^2(b+c)^2a(b^3+c^3-a^3)\geq$$
$$\geq a^2(b-c)^2(a+c)^2b(a^3-b^3)+b^2(b-c)^2(b+c)^2a(b^3-a^3)=$$
$$=ab(b-c)^2(a^3-b^3)(a(a+c)^2-b(b+c)^2)\geq0.$$
Also, $$b^2\sum_{cyc}(a-b)^2(a+b)^3c^2(a+b-c)\geq$$
$$\geq b^2(a-c)^2(a+c)^3b^2(a+c-b)+b^2(b-c)^2(b+c)^3a^2(b+c-a)\geq$$
$$\geq a^2(b-c)^2(a+c)^3b^2(a-b)+b^2(b-c)^2(b+c)^3a^2(b-a)=$$
$$=a^2b^2(b-c)^2(a-b)((a+c)^3-(b+c)^3)\geq0$$ and we are done! 
A: Also, we can use SOS after the following C-S:
$$\sum_{cyc}\frac{a^2}{(b+c)^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}$$ and it's remains to prove that
$$\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}+\frac{1}{4}\geq\frac{a^2+b^2+c^2}{ab+ac+bc},$$ which is sixth degree already and it's obvious by $uvw$, which you don't want.
The last inequality we can prove also by the following reasoning.
Since $$(ab+ac+bc)\sum_{cyc}(a^2+ab)\geq3\sum_{cyc}(a^2b^2+a^2bc)$$ it's $$\sum_{cyc}ab(a-b)^2\geq0,$$ we obtain:
$$\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}a^2(b+c)^2}=\frac{(a^2+b^2+c^2)^2}{2\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{(a^2+b^2+c^2)^2}{\frac{2}{3}(ab+ac+bc)\sum\limits_{cyc}(a^2+ab)}.$$
Thus, it's enough to prove that
$$\frac{(a^2+b^2+c^2)^2}{\frac{2}{3}(ab+ac+bc)\sum\limits_{cyc}(a^2+ab)}+\frac{1}{4}\geq\frac{a^2+b^2+c^2}{ab+ac+bc}.$$
Now, let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, $k\geq1$ and we need to prove that
$$\frac{k^2}{\frac{2}{3}(k+1)}+\frac{1}{4}\geq k$$ or
$$(k-1)(2k-1)\geq0$$ and we are done!
