Find bound for sum of square roots
Solution 1:
Assuming $a_i\geq 0$, you could consider the vector $x=(\sqrt{a_1},...,\sqrt{a_n})$, then you have $||x||_2^2 = A$, and Hölder's inequality gives $||x||_1 \leq \sqrt{n} ||x||_2$, from which you get the bound
$\sqrt{a_1}+...+\sqrt{a_n} \leq \sqrt{n A}$
Solution 2:
Hint: Use the Cauchy-Schwarz Inequality. We have $$\left(\sum\sqrt{a_i}\right)^2\le n\sum a_i.$$ (In the notation of the article linked to, $x_i=\sqrt{a_i}$ and $y_i=1$.)