Ta có:
\(\dfrac{n}{a+b+c}=\dfrac{100a+10b+c}{a+b+c}=1+\dfrac{99a+9b}{a+b+c}\)
\(\ge1+\dfrac{99a+9b}{a+b+9}=10+\dfrac{90a-81}{a+b+9}\ge10+\dfrac{90a-81}{a+18}\)
\(=100+\dfrac{-1701}{a+18}\ge100-\dfrac{1701}{19}=\dfrac{199}{19}\)
Dấu = xảy ra khi:\(\left\{{}\begin{matrix}a=1\\b=c=9\end{matrix}\right.\)