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


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 $abc=ab\ge25$.
