Áp dụng bất đẳng thức Cosi, ta có :
\(53=2x+3y\ge2\sqrt{2x.3y}=2\sqrt{6}.\sqrt{xy}\Rightarrow xy\le\left(\frac{53}{2\sqrt{6}}\right)^2\)
Do đó : \(P=\sqrt{xy+4}\le\sqrt{\left(\frac{53}{2\sqrt{6}}\right)^2+4}=\sqrt{\frac{2905}{24}}\)
Vậy : Max \(P=\sqrt{\frac{2905}{24}}\Leftrightarrow\left(x;y\right)=\left(\frac{53}{4};\frac{53}{6}\right)\)