$\begin{array}{l} P = \left[ {\left( {\dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt y }}} \right).\dfrac{2}{{\sqrt x + \sqrt y }} + \dfrac{1}{x} + \dfrac{1}{y}} \right]:\dfrac{{\sqrt {{x^3}} + y\sqrt x + x\sqrt y + \sqrt {{y^3}} }}{{\sqrt {{x^3}y} + \sqrt {x{y^3}} }}\\ P = \left[ {\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}.\dfrac{2}{{\sqrt x + \sqrt y }} + \dfrac{{x + y}}{{xy}}} \right].\dfrac{{\sqrt {xy} \left( {x + y} \right)}}{{x\sqrt x + y\sqrt x + x\sqrt y + y\sqrt y }}\\ P = \left( {\dfrac{2}{{\sqrt {xy} }} + \dfrac{{x + y}}{{xy}}} \right).\dfrac{{\sqrt {xy} \left( {x + y} \right)}}{{\left( {x + y} \right)\left( {\sqrt x + \sqrt y } \right)}}\\ P = \dfrac{{2\sqrt {xy} + x + y}}{{xy}}.\dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }} = \dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{xy}}.\dfrac{{\sqrt {xy} }}{{\sqrt x + \sqrt y }}\\ P = \dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }}\\ b)\dfrac{{\sqrt x + \sqrt y }}{{\sqrt {xy} }} = \dfrac{{\sqrt x + \sqrt y }}{{\sqrt {16} }} = \dfrac{{\sqrt x + \sqrt y }}{4} \ge \dfrac{{2\sqrt {\sqrt {xy} } }}{4} = \dfrac{{2.\sqrt {\sqrt {16} } }}{4} = \dfrac{{2.\sqrt 4 }}{4} = 1\\ \Rightarrow \min P = 1\\ "=" \to \left\{ \begin{array}{l} \sqrt x = \sqrt y \\ xy = 16 \end{array} \right. \Rightarrow x = y = 4 \end{array}$
