1708\hspace{-16}$Here $\mathbf{\alpha\;,\beta}$ are two distinct Roots of the equation. $\mathbf{ax^2-bx+c=0}$\\\\ Then $\mathbf{\alpha+\beta =\frac{b}{a}}$ and $\mathbf{\alpha.\beta=\frac{c}{a}}\;\;,$ Where $\mathbf{0<\alpha,\beta<1\;, a,b,c \in \mathbb{N}}$\\\\ Now $\mathbf{0<(1-\alpha),(1-\beta)<1}$\\\\ So Using $\mathbf{A.M\geq G.M}$\\\\ $\mathbf{\frac{\alpha+(1-\alpha)}{2}\geq \sqrt{\alpha.(1-\alpha)}\Leftrightarrow \alpha.(1-\alpha)\leq \frac{1}{4}}$\\\\ Similarly $\mathbf{\beta.(1-\beta)\leq \frac{1}{4}}$\\\\ So $\mathbf{\alpha.\beta.(1-\alpha).(1-\beta)\leq \frac{1}{16}}$\\\\ $\mathbf{\alpha.\beta.\{\alpha.\beta -(\alpha+\beta)+1\}\leq \frac{1}{16}}$\\\\ So $\mathbf{c.(c-b+a)\leq \frac{a^2}{16}}$\\\\ So $\mathbf{a^2\geq 16\Leftrightarrow a\geq 5}$\\\\ bcz Here $\mathbf{a,b,c\in\mathbb{N}}$ and $\mathbf{\min(c.(c-b+a))=1}$\\\\
\hspace{-16}$Now for Distinct Roots, $\mathbf{D>0\Leftrightarrow b^2-4ac>0>20a\geq 20}$\\\\ So $\mathbf{b\geq 5}$ and $\mathbf{c\geq 1}$\\\\ Now $\mathbf{a,b,c\in \mathbb{N}}$\\\\ So $\mathbf{a_{min}\;,b_{min}=5}$ and $\mathbf{c_{min}=1}$\\\\ So $\mathbf{a.b.c\geq 5.5.1\Leftrightarrow \log_{5}(a.b.c)\geq \log_{5}5^2}$\\\\ So $\mathbf{\log_{5}(a.b.c)\geq 2}$