Polynomial inequality of sixth degree There is the following problem.

Let $a$, $b$ and $c$ be real numbers such that $\prod\limits_{cyc}(a+b)\neq0$ and $k\geq2$ such that $\sum\limits_{cyc}(a^2+kab)\geq0.$ 
  Prove that:
  $$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}\geq\frac{9}{4}.$$

I have a proof of this inequality for any $k\geq2.6$.
I think, for $k<2.6$ it's wrong, but my software does not give me a counterexample 
and I don't know, how to prove it for some $k<2.6$.    
It's interesting that without condition $\sum\limits_{cyc}(a^2+kab)\geq0$ the equality occurs also for $(a,b,c)=(1,1,-1)$. 
My question is:  What is a minimal value of $k$, for which this inequality is true?
Thank you! 
 A: Your inequality fails to hold when e.g. $k=25999/10000=2.6-10^{-4}$ and $(a,b,c)=(97661/65536,-5/3,-1)$.

Indeed, the smallest value for which your inequality holds is $13/5=2.6$. Here is a proof by Mathematica: 


So, the value $13/5$ of $k$ is witnessed by 
$$a=-1,\ b=x_*,\ c=\frac{1}{10} \left(13-13 x_*-\sqrt{69 x_*^2-78
   x_*+69}\right),\tag{1}$$
where $x_*=-1.68\ldots$ is the smallest root of the $6$ real roots of the polynomial 
$$p(x)=1681 - 3198 x - 3621 x^2 + 10292 x^3 - 3621 x^4 - 3198 x^5 + 1681 x^6.$$
For this proof, Mathematica took about 32 sec, which is a huge time for a computer. 

The values of the sums 
$$(s_1,s_2,s_3):=\left(\frac{2 a^2+b c}{(b+c)^2}+\frac{2b^2+a c}{(a+c)^2}+\frac{2 c^2+a b}{(a+b)^2},a^2+b^2+c^2,a b+a c+b
   c\right)$$
for the extremal $(a,b,c)$ given by (1) are
$$\Big(\frac94,-\frac{13}5\,s_{3*},s_{3*}\Big),$$
where $s_{3*}=-2.34\ldots$ is the smallest root of the $3$ real roots of the polynomial 
$$p_3(x)=3375 + 8775 x + 7065 x^2 + 1681 x^3,$$
with the other two roots $-1.04\ldots$ and $-0.826\ldots$. 
A: We want to show that your inequality does not hold for $k\in[2,13/5)$. In view of the identity in your answer, it is enough to show that for each $k\in[2,13/5)$ there is a triple $(a,b,c)\in\mathbb R^3$ with the following properties: $a=-1>b$, 
\begin{align}s_4&:=\sum_{cyc}(2a^3-a^2b-a^2c) \\
&=a^2 (2 a-b-c)+b^2 (-a+2 b-c)+c^2 (-a-b+2 c)=0,
\end{align}
$$s_3:=a b + b c + c a<0,$$
and 
$$s_2+k s_3=0,$$
where
$$s_2:=a^2 + b^2 + c^2.$$
Indeed, then the right-hand side of your identity will be 
$$\frac{20}{3}\sum_{cyc}(a^4-a^2b^2)(13/5-k)s_3<0,$$
so that your identity will yield
$$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}<9/4.$$
For each $k\in(2,13/5)$, the triple $(a,b,c)$ will have all the mentioned properties if $a=-1$, $b$ is the smallest (say) of the 6 real roots $x$ of the polynomial 
$$P_k(x):=-18 - 15 k + 4 k^2 + 
 4 k^3 + (36 k + 6 k^2 - 12 k^3) x + (-27 - 9 k - 21 k^2 + 
    6 k^3) x^2 + (18 + 60 k - 10 k^2 + 8 k^3) x^3 + (-27 - 9 k - 
    21 k^2 + 6 k^3) x^4 + (36 k + 6 k^2 - 12 k^3) x^5 + (-18 - 15 k + 
    4 k^2 + 4 k^3) x^6, $$
and 
$$c=\tfrac12\, (k - b k) - \tfrac12\, \sqrt{-4 - 4 b^2 + 4 b k + k^2 - 2 b k^2 + b^2 k^2}.$$
For $k=2$, $(a,b,c)=(-1,0,1)$ will be such a triple. 
So, we are done.

This result was obtained with Mathematica, as follows (which took Mathematica about 0.05 sec):

