Solution:
Let $x$ and $y$ be the roots. Then:

$$\begin{gathered} \frac{b}{a}=x+y<2 \Rightarrow b<2a \\ \frac{c}{a}=x y<1 \Rightarrow c<a \Rightarrow a>1 \\ b^{2}>4 a c>4 c^{2} \Rightarrow b>2c \end{gathered}$$

Evaluated at $1$, the polynomial must be greater than $0$, so $a+c>b$. Then:

$$\begin{gathered} 2c < b < a+c \\ 2c+1 \leq b \leq a+c-1 \\ a \geq c+2 \geq 3 \end{gathered}$$

If $a=3$, then $c=1$ and $b=3$, by the above bounds, but this polynomial has complex roots. Similarly, if $a=4$, then $c=1$ and $b$ is forced to be either $3$ or $4$, again giving either $0$ or $1$ distinct real roots. So $a \geq 5$. But the polynomial $5 x^{2}-5 x+1$ satisfies the condition, so $5$ is the answer.