If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ $a,b,c$ belongs to natural prove that $\log_5 {abc}\geq2$

If $ax^2-bx+c=0$ has two distinct real roots lying in the interval $(0,1)$ with $a, b, c\in \mathbb N$, prove that $\log_5 {abc}\geq2$.

The equations I could form are:

1) $f(0)>0$ and $f(1)>0$

2) $\frac{b}{2a}$ lies between $0$ and $1$, because: $\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}<\frac{b}{2a}<\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a}$.

3) $\Delta=b^2-4ac > 0$

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What is $d$? What does $\frac{b}{2a}$ lie between? @maths lover, fix your question. – Soham Chowdhury Jul 20 '13 at 17:36
@SohamChowdhury: now how does it look? – The Chaz 2.0 Jul 20 '13 at 20:09
For consistency of the problem assume $a>$ 0. If $c =0$ or $b = 0$ does not have the roots of $f (x) = ax^2-bx+c$ in $(0,1)$. Therefore $c\geq 1$ and $b\geq 1$.$\log_5(abc)>2 \Longleftrightarrow \log_5(abc)>\log_5(5^2) \Longleftrightarrow abc>25.$ – MathOverview Jul 20 '13 at 22:29

We need to show that $abc\ge25$. Since both roots are real and distinct, we have that $b^2-4ac>0$ and so $b^2>4ac$. Since both roots are in (0,1), their average $\frac{b}{2a}<1$ and therefore $b<2a$. Since the larger root $\frac{b+\sqrt{b^2-4ac}}{2a}<1$, we have that $b+\sqrt{b^2-4ac}<2a$ and therefore $\sqrt{b^2-4ac}<2a-b$. Squaring both sides gives $b^2-4ac<4a^2-4ab+b^2$, so $4ab<4a^2+4ac$ and therefore $b<a+c$. Since $2a>b$, we have that $4a^2>b^2>4ac$ and therefore $a>c$. Since $b^2\ge4ac+1$ and $a\ge c+1$, we conclude that $b^2\ge 4c(c+1)+1=(2c+1)^2$ and thus $b\ge 2c+1$. Therefore $abc\ge (c+1)(2c+1)c>25$ if $c\ge 2$. When $c=1$, $b<a+1$ implies that $b\le a$, so $a^2\ge b^2>4a$ and therefore $a\ge 5$. Then $b^2>4a\ge20$, so $b\ge5$ and $abc=ab\ge25$.
Where did you get $b^2 \ge 4ac+1$ from? – Ovi Jul 21 '13 at 0:14
I was using $b^2>4ac$ – user84413 Jul 21 '13 at 17:58
But how does $b^2>4ac$ imply $b^2 \ge 4ac+1$ ? – Ovi Jul 21 '13 at 18:15