Logarithmic quadratic inequality

Given that the roots must lie in $(0,1)$ . The roots will be $\frac{b \pm \sqrt{b^2-4ac}}{2a}$ . As $a,b,c$ are natural so clearly $\frac{b \pm \sqrt{b^2-4ac}}{2a}>0$ . This will remain true for all $a,b,c$ .

Now for the other condition: $\frac{b \pm \sqrt{b^2-4ac}}{2a}<1$

Let us first take, $\frac{b + \sqrt{b^2-4ac}}{2a}<1\Rightarrow b <a+c \ldots$ A

Secondly take, $\frac{b -\sqrt{b^2-4ac}}{2a}<1$

We will have to take two cases for it:

Case 1 $b-2a>0$

Putting this would give us $b>a+c$ but this contradicts A .Hence, we blindly accept Case 2 .

Case 2 $b-2a \leq 0$

Hence we have another inequality as $b-2a \leq 0 \ldots$ B

Also as the roots are distinct so $b^2>4ac \ldots$ C

Try putting C in A to get $a\neq c \ldots$ D

Now we focus on C . Try first $b=1$ then $b=2$ and so on and every supposition will contradict any of A , B or D . The least value you will get by putting $b=5$ to give $a=6$ and $c=1$

This would give us $\lambda = 2$ .

@Sandeep Bhardwaj Sir,please post your solution(I got it correct, but still want to see the best and shortest way to do it)