Find bound for sum of square roots Let $a_1,...,a_n$ be real numbers, such that $a_1+...+a_n=A$.
What can we say about $\sqrt{a_1}+...+\sqrt{a_n}$?
I would like to bound from above thus sum in terms of $A$.
 A: Since we are taking the square root of each $a_i$, we will assume that $a_i\ge0$.
Using Jensen's Inequality yields
$$
\left(\frac1n\sum_{i=1}^n\sqrt{a_i}\right)^2\le\frac1n\sum_{i=1}^n\left(\sqrt{a_i}\right)^2=\frac1n\sum_{i=1}^na_i\tag{1}
$$
Rearranging $(1)$ gives
$$
\sum_{i=1}^n\sqrt{a_i}\le\sqrt{n\sum_{i=1}^na_i}\tag{2}
$$
A: Just for fun, you can also look at it in terms of elementary statistics. For $k=1,\dots,n$ let $x_k=\sqrt{a_k}$. Let $\bar x$ be the mean of the $x_k$. For fixed $\bar x$, the variance of the $x_k$ is minimized when $x_1=\ldots=x_n$, when it is $0$. But the variance is $\frac1n\sum_k x_k^2-\bar x^2$, so the minimum value of $\sum_k x_k^2$ is $n\bar x^2$, occurring when $x_1=\ldots=x_n$. For fixed $A=\sum_k x_k^2$, therefore, the maximum value of $\bar x$ is $\sqrt{A/n}$, and hence $\sum_k x_k\le \sqrt{nA}$, with equality when $x_1=\ldots=x_n$.
A: 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
$\sqrt{a_1}+...+\sqrt{a_n} \leq \sqrt{n A}$
A: 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 $y_i=1$.)
