Max. value $M$ of $c^2$ for which points lie opposite of $x(a+b)^2-(ab+bc+ca+1)y+2 = 0$ Let $M$ be the the maximum value of $c^2$ for which $O(0,0)$ and $A(1,1)$ does not lie on 
opposite side of straight line $x(a+b)^2 -(ab+bc+ca+1)y +2=0.$Then value of $M$ is
$\bf{My\; Solution::}$ Given $O(0,0)$ and $A(1,1)$ does not lie on opposite side of line $L_{(x,y)}$ means
These two points lie on same side of line. So $L_{(0,0)}\times L_{(1,1)}>0$
So here $L_{(x,y)} = x(a+b)^2 -(ab+bc+ca+1)y +2.$
So $L_{(0,0)} = 2$ and $L_{(1,1)} = (a+b)^2-(ab+bc+ca+1)+2$
So $(a+b)^2-(ab+bc+ca+1)+2>0$
Now How can I solve after that
Help Required
Thanks

Hint:
\begin{align}
(a+b)^2-ab -bc - ca +1 &\ge (a+b)^2 - \tfrac12(a^2+b^2) - \tfrac16(9b^2+c^2)-\tfrac16 (c^2+9a^2)+1 \\
&= -(a-b)^2+1 - \tfrac13 c^2 \\
&\ge 1- \tfrac13 c^2
\end{align}
