Let's say $a_1, a_2, ..., a_n$ are positive real numbers and $a_1 + a_2 + ... + a_n = 1$

I've to prove the following expression using the Cauchy-Schwarz inequality but I don't know how to do it.

$\sqrt{{a_1}} + \sqrt{{a_2}} + \dots + \sqrt{{a_n}} \leq \sqrt{n}$

Choosing a second set of real numbers $b_1 = b_2 = \dots b_n = 1$ and applying Cauchy-Schwarz inequality, I got the next inequality, which is almost trivial:

$ 1 \leq \sqrt{n} . \sqrt{{a_1^2}+{a_2^2}+\dots+{a_n^2}}$

but I think is a dead end and isn't the correct way to prove it.

Please, any ideas?

Thanks so much in advance.


2 Answers 2


Let $\textbf{a}=(\sqrt{a_1},\sqrt{a_2},\dots,\sqrt{a_n})$ and $\textbf{b}=(1,1,\dots,1)$. Then $\|\textbf{a}\|=1$ and $\|\textbf{b}\|=\sqrt{n}$. Thus $$\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_n}=\langle \textbf{a},\textbf{b} \rangle \leq \|\textbf{a}\| \|\textbf{b}\|=\sqrt{n}$$

  • 1
    $\begingroup$ Perfect and really simple. Makes me feel as a stupid... :| Thanks so much. $\endgroup$
    – cooper
    Oct 22, 2019 at 1:00

$\sum c_k b_k \leq \sqrt {\sum c_k^{2}}\sqrt {\sum b_k^{2}}$. Put $c_k=\sqrt a_k$ and $b_k=1$.

  • $\begingroup$ @fordjones No. $\sum b_k^2 = n$. $\endgroup$
    – amsmath
    Oct 22, 2019 at 0:55
  • $\begingroup$ @fordjones I think is correct. The first sum is equal to "n" and the second one is equal to 1. $\endgroup$
    – cooper
    Oct 22, 2019 at 0:59
  • $\begingroup$ never mind made a dumb calculation error again +1 $\endgroup$
    – user706791
    Oct 22, 2019 at 2:00

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