KöMaL 2016 A681

The positive numbers $a_1,a_2,\ldots,a_n$ satisfy $\sqrt{a_1}+\sqrt{a_2} +\cdots+\sqrt{a_n}=1$. Show that ${a_1}^{a_1}\cdot {a_2}^{a_2}\cdot \ldots \cdot {a_n}^{a_n} \ge {(a_1+a_2+\ldots+a_n)}^2.$