# Proof using Cauchy-Schwarz inequality

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.

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 $$\sqrt{a_1}+\sqrt{a_2}+\cdots+\sqrt{a_n}=\langle \textbf{a},\textbf{b} \rangle \leq \|\textbf{a}\| \|\textbf{b}\|=\sqrt{n}$$
$$\sum c_k b_k \leq \sqrt {\sum c_k^{2}}\sqrt {\sum b_k^{2}}$$. Put $$c_k=\sqrt a_k$$ and $$b_k=1$$.
• @fordjones No. $\sum b_k^2 = n$. Oct 22, 2019 at 0:55