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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 $A$.

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$a_i$ are all nonnegative? – Paul Apr 12 '12 at 6:26
Yes, we can assume that they are nonnegative. – Nick G.H. Apr 12 '12 at 6:48
up vote 15 down vote accepted

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

$\sqrt{a_1}+...+\sqrt{a_n} \leq \sqrt{n A}$

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This bound is 'tight' in the sense that if you take all $a_i$ equal, then it is an equality. – copper.hat Apr 12 '12 at 6:39

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 $y_i=1$.)

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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 $$ \sum_{i=1}^n\sqrt{a_i}\le\sqrt{n\sum_{i=1}^na_i}\tag{2} $$

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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 $x_1=\ldots=x_n$.

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