\(P=\dfrac{100a+10b+c}{a+b+c}\le\dfrac{100a+100b+100c}{a+b+c}=100\)
\(P_{max}=100\) khi \(b=c=0\)
Mặt khác ta có \(\left\{{}\begin{matrix}a\ge1\\c\le9\end{matrix}\right.\) \(\Rightarrow9a\ge c\Rightarrow90a\ge10c>9c\)
\(\Rightarrow P=\dfrac{10a+90a+10b+c}{a+b+c}>\dfrac{10a+9c+10b+c}{a+b+c}=10\)
Hay \(P-10>0\)
Ta cần tìm số k lớn nhất sao cho: \(\dfrac{100a+10b+c}{a+b+c}\ge k\) đồng thời \(10< k\le100\)
\(\Leftrightarrow100a+10b+c\ge ka+kb+kc\)
\(\Leftrightarrow\left(100-k\right)a\ge\left(k-10\right)b+\left(k-1\right)c\)
Mà \(\left\{{}\begin{matrix}\left(100-k\right)a\ge100-k\\\left(k-10\right)b+\left(k-1\right)c\le\left(k-10\right).9+\left(k-1\right).9=18k-99\end{matrix}\right.\)
\(\Rightarrow100-k\ge18k-99\Rightarrow k\le\dfrac{199}{19}\)
\(\Rightarrow k=\dfrac{199}{19}\)
Hay \(P_{min}=\dfrac{199}{19}\) khi \(\overline{abc}=199\)
