An algebra problem by Archit Tripathi

Call the supremum sought $S$ . If $a=0$ , then the inequality reduces to $x>\frac{1}{3}$ , which is satisfied for many positive $x$ . This establishes $a=0$ as a lower bound for S. Now I will consider $a>0$ . In order to have a positive solution, two conditions must be satisfied: the discriminant of ax^2+(a-3)x+1 must be positive and the larger root of $ax^2+(a-3)x+1=0$ must be positive (also note that the converse is true as well). This leads to the pair of inequalities $(a-1)(a-9)>0$ and $\sqrt{(a-1)(a-9)}>a-3$ . The first inequality is equivalent to $a<1$ or $a>9$ (otherwise the parabola is strictly above the $x$ -axis). First consider $a>9$ . Since both sides of $\sqrt{(a-1)(a-9)}>a-3$ are positive, I can square both sides to obtain $a^2-10a+9>a^2-6a+9\Rightarrow a<0$ . This is contradictory, so now I consider $a<1$ . For this case, it is easy to see that, in $\sqrt{(a-1)(a-9)}>a-3$ , the LHS is positive, and the RHS is negative. Therefore, for all $a\in (0,1)$ , the inequality has a positive solution. And recall that for all $a>1$ , the inequality has no positive solution. Hence $S=1$ .