$\begin{align} & {{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}>0 \\ & a+b>2\sqrt{ab} \\ & \text{Also} \\ & {{\left( \frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}} \right)}^{2}}>0 \\ & \frac{1}{a}+\frac{1}{b}-\frac{2}{\sqrt{ab}}>0 \\ & \frac{1}{a}+\frac{1}{b}>\frac{2}{\sqrt{ab}} \\ & \therefore \left( a+b \right)\left( \frac{1}{a}+\frac{1}{b} \right)>2\sqrt{ab}\times \frac{2}{\sqrt{ab}} \\ & \left( a+b \right)\left( \frac{1}{a}+\frac{1}{b} \right)>4 \\\end{align}$