What is the minimum value of $abc$

Question

If the roots of the equation $$ax^2-bx+c=0$$ lie in the interval $(0,1)$, find the minimum possible value of $abc$.

Edit: I forgot to mention in the question that $a$, $b$, and $c$ are natural numbers. Sorry for the inconvenience.
Edit 2: As Hagen von Eitzen said about the double roots not allowed, I forgot to mention that too. Extremely sorry :(

I tried to use $D > 0$, where $D$ is the discriminant but I can't further analyze in terms of the coefficients. Thanks in advance!

Answer 1

Given: Roots lie in $(0,1).$

Let $f(x)=ax^2-bx+c$ and it's roots be $\alpha$ and $\beta$

$\implies f(0) \times f(1) > 0$ (Can be easily verified from the parabolic graph of $f(x)$)

or $c(a-b+c)>0$

$\implies \frac{c}{a}(1-\frac{b}{a}+\frac{c}{a})>0$

$\implies \alpha \beta (1-(\alpha + \beta) + \alpha\beta) > 0$ (Using Vieta's Formula)

$\implies \alpha\beta(1-\alpha)(1-\beta) > 0$

Now,

Consider $\alpha(1-\alpha)$,

By AM-GM inequality,

$\alpha(1-\alpha)<\frac{1}{4}$

Similarly,

$\beta(1-\beta)< \frac{1}{4}$

By multiplying the above two inequalities, we get,

$\alpha\beta(1-\alpha)(1-\beta)<\frac{1}{16}$

$\implies \frac{c}{a}(1-\frac{b}{a}+\frac{c}{a})<\frac{1}{16}$

$\implies c(a-b+c)<\frac{a^2}{16}$

If we let $a$=$b$ and $c=1$, clearly we are getting the minimum value of $a$, i.e.

$\frac{a^2}{16}>1$

$a>4$ or minimum $a =5$

Since $D > 0$, we have $b^2-4ac > 0$ (where $D$ is the discriminant of $f(x)=0$)

this inequality is satisfied for $a=b=5$ which we calculated above

thus at $a=b=5$ and $c=1$ the minimum value of $abc=25$ is achieved.

Answer 2

If you multiply the equation by $k$, you get $$(ka)x^2-(bk)x+(ck)=0$$ This new equation has the same roots as the original, hence in $(0,1)$, but has the product of its coefficients $k^3abc$. By letting $k\to\pm \infty$ (depending on whether $abc>0$ or $abc<0$), you can make this product as small as you like. Hence the answer is $-\infty$.

Answer 3

The answer is $a=4$, $b=4$, $c=1$, giving $x = \frac12$ (twice), and a product $abc=16$. Exhaustive search through all $1 \leq a,b,c \leq 16$ gave no better answer.

Answer 4

The discrimimnat $D=b^2-4ac$ must be positive to ensure two distinct real roots. (If double root is not forbidden, we have $4x^2-4x+1$ with double root at $\frac12$ and $abc=16$). Next, we must have $f(1)>0$, i.e. $$a+c>b.$$ For naturals $a,c$ we also have $ a+c\le 1+ac$ and conclude $$\tag1b\le ac.$$ If $b\le 4$ we obtain $b\le ac<\frac14b^2\le b$, contradiction. (NB: If we relax the condition that the roots be distinct, the $<$ becomes a $\le$ and instead of a contradiction we find $b=ac=4$, hence $abc=16$). Hence $b\ge 5$ and by $(1)$ $$abc\ge b^2\ge 25.$$ The minimum is indeed attained as can be seen by making all iniequalities sharp, which gives: Either $(a,b,c)=(5,5,1)$ or $(a,b,c)=(1,5,5)$. The first of these indeed gives two roots in $(0,1)$.