\(\left(x+y\right)^2\ge4xy=4.16=64\Rightarrow x+y\ge8\)
\(P=x+y+\dfrac{4}{x+y}\)
\(=\left(\dfrac{x+y}{16}+\dfrac{4}{x+y}\right)+\dfrac{15\left(x+y\right)}{16}\)
\(\ge2\sqrt{\dfrac{x+y}{16}.\dfrac{4}{x+y}}+\dfrac{15.8}{16}\)
\(=2.\dfrac{1}{2}+\dfrac{15}{2}=\dfrac{17}{2}\)
- Vậy \(MinP=\dfrac{17}{2}\), đạt tại \(x=y=4\)