Có \(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\)
\(\Leftrightarrow\frac{1}{2}\ge\frac{2}{\sqrt{xy}}\Leftrightarrow\sqrt{xy}\ge4\)
P=\(\sqrt{x}+\sqrt{y}\) \(\Rightarrow\) \(P^2=\left(\sqrt{x}+\sqrt{y}\right)^2\ge4\sqrt{xy}\Leftrightarrow\ge4.4=16\)
\(\Rightarrow P\ge4\)