\(P=\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{\left(1+1+2\right)^2}{a+b+c}=4\)
Suy ra \(minP=4\).
Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}=\frac{2}{c}\\a+b+c=4\\a,b,c>0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b=1\\c=2\end{cases}}\).