Let $a_{n+1}=b_n+c_n, b_{n+1}=a_n+c_n$ and $c_{n+1}=a_n+b_n$ for all $n\ge 1$. We are trying to find a positive integer $k$ such that the equation $a_kx^2+b_kx+c_k=0$ doesn't have real roots. For the sake of contradiction, suppose that the equation $a_nx^2+b_nx+c_n=0$ has $2$ real roots for all $n\ge 1$. We will use the condition $(a_1+c_1)b_1>0$ just to prove that $a_1+b_1+c_1\neq 0$, which is quite obvious. If $a_1+b_1+c_1=0$, then $a_1+c_1=-b_1$. Thus, $0<(a_1+c_1)b_1=-b_1^2\Rightarrow b_1^2<0$. Contradiction. By induction over $n$, we can easily prove that $a_n=T_n+(-1)^{n-1}a_1$ , $b_n=T_n+(-1)^{n-1}b_1$ and $c_n=T_n+(-1)^{n-1}c_1$ where $T_n=(a_1+b_1+c_1)\cdot \dfrac{2^{n-1}+(-1)^n}{3}$ for all $n\ge 1$. Since $a_1+b_1+c_1\neq 0$ and $\dfrac{2^{n-1}+(-1)^n}{3}$ is unbounded, we have $T_{n}$ is unbounded and strictly increasing (or decreasing). We know that $b_{2t}\ge 4a_{2t}\cdot c_{2t}\Rightarrow (T_{2t}-b_1)^2\ge 4(T_{2t}-a_1)(T_{2t}-c_1)\Rightarrow 3T_{2t}^2+T_{2t}(2b_1-4a_1-4c_1)+(4a_1c_1-b_1^2)\leq 0$. The numbers $2b_1-4a_1-4c_1$ and $4a_1c_1-b_1^2$ are fixed. Also, $3>0$. Then, the function $f(t)=3T_{2t}^2+(2b_1-4a_1-4c_1)+(a_1\cdot c_1-b_1^2)$ is greater than $0$ for some $t$. Contradiction.