Sum of Square-Weights For positive reals $a,b,c$, prove that $$\frac{a^3+b^3+c^3}{3abc}+\sum_{\text{cyc}} \frac{a(b+c)}{b^2+c^2}\geq 4$$
I've heard of a lemma stating if a polynomial expression $f(a,b,c)$ satisfies both 
$f(a,b,c)=f(b,a,c)$ and $f(a,a,c)=0$, then it is divisable by $(a-b)^2$. I tried splitting up the expression into different polynomials where the lemma could be applied but had no success. 
I would appreciate any help.
Thanks in advance.

$\sum_{\text{cyc}} \dfrac{a(b+c)}{b^2+c^2} -3=\sum_{\text{cyc}}(\dfrac{a(b+c)}{b^2+c^2} -1 )=\sum_{\text{cyc}}\dfrac{b(a-b)+c(a-c)}{b^2+c^2}=\sum_{\text{cyc}}(\dfrac{b(a-b)+}{b^2+c^2}+\dfrac{a(b-a)}{a^2+c^2})=\sum_{\text{cyc}}(\dfrac{(a-b)^2(ab-c^2)}{(b^2+c^2)(a^2+c^2)})=\sum \dfrac{(a-b)^2(ab)}{(b^2+c^2)(a^2+c^2)}-\sum \dfrac{(a-b)^2c^2}{(b^2+c^2)(a^2+c^2)}\ge \sum \dfrac{(a-b)^2(ab)}{(b^2+c^2)(a^2+c^2)}-\sum \dfrac{(a-b)^2c}{4abc}$
$\dfrac{a^3+b^3+c^3}{3abc}-1 =\sum \dfrac{(a+b+c)(a-b)^2}{6abc}$
$\sum \dfrac{(a+b+c)(a-b)^2}{6abc} -\sum \dfrac{(a-b)^2c}{4abc}=\dfrac{1}{12abc}\sum (a-b)^2(2(a+b+c)-3c)=\dfrac{1}{12abc}\sum (a-b)^2(a+b)+\dfrac{1}{12abc}\sum (a-b)^2(a+b-c)$
$\sum (a-b)^2(a+b-c)\ge 0 \iff a^3+b^3+c^3+3abc \ge \sum a^2b$
