Find minimal value of $abc$ if the quadratic equation $ax^2-bx+c = 0$ has two roots in $(0,1)$

Question

If $$ ax^2-bx+c = 0 $$ has two distinct real roots in (0,1) where a, b, c are natural numbers then find the minimum value of product abc ?

Answer 1

Hints: If the roots are $\;\alpha,\,\beta\;$ , then

$$\alpha+\beta=\frac ba\;,\;\;\alpha\beta=\frac ca\,,\,\,\text{and we can also write}\;\; abc=a^3\cdot\frac ba\cdot\frac ca\;\ldots$$

Answer 2

Since $a,b,c $ are positive the roots are trivially greater than 0.

What remains is to solve the inequality:

$\frac{b + \sqrt{b^2-4ac}}{2a} <1$

This reduces to $ a+c>b$

But the roots being real and distinct we have $b^2 >4ac$

Combining both we have :

$a^2 + c^2 + 2ac > b^2 > 4ac$

$b^2 > 4ac$ tells us $b> 2$ (why?)

$ a^2 + c^2+ 2ac > 4ac $ tells us $a \neq c$

Checking small cases we get $(a,b,c) =(5,5,1)$ where $abc =25$

EDIT:

Checking "small" cases is not informative, so adding an explanation:

Keeping in mind $a+c>b$, the minimum value of $ac$ occurs when $a=b$ and $c=1$. So for given $b$, the minim of $abc$ is $b^2$. The smallest value of $b$ which agrees the inequality $b^2>4b$ is 5 (as $ac=b$). Hence the corresponding minimum value is $5^2$