First, by AGM, we have $$a+b \ge 2\sqrt{ab}$$ Next, apply AGM again, we have \begin{align} \sqrt{\frac{a+b}{2}}&=\sqrt{\frac{(\sqrt{a})^2+(\sqrt{b})^2}{2}}\\ &=\sqrt{\frac{((\sqrt{a})^2+(\sqrt{b})^2)+((\sqrt{a})^2+(\sqrt{b})^2)}{4}}\\ &\ge\sqrt{\frac{(\sqrt{a})^2+(\sqrt{b})^2+2(\sqrt{a})(\sqrt{b})}{4}}\\ &= \frac{\sqrt{a}+\sqrt{b}}{2} \end{align} Multiplying these two relations yields the result.