Inner Product Spaces Inequality Question

Question

Prove that for any natural number $n$ and real numbers $x_1, x_2, ..., x_n$ we have the inequality $$\left|\dfrac{1}{n}\sum_{k=1}^nx_k\right| \le\left(\dfrac{1}{n}\sum_{k=1}^nx^2_k\right)^\dfrac{1}{2}$$

My first thought was using the Cauchy-Schwarz Inequality but I am not sure if we can use only one variable in it and

(edit so the next part of the sentence is probably not relevant but I will keep it anyway) that wouldn't include the $\dfrac{1}{n}$ on the left hand side and $\dfrac{1}{2}$ on the right hand side.

The next thing I was looking at was the norm of x. $$\|\textbf{x}\|=\sqrt{\textbf{x}\cdot \textbf{x}}=\sqrt{\sum_{k=1}^nx^2_k}$$ And $$\|\textbf{x}\|_1=\sum_{k=1}^n|x_k|$$ But I don't see how that would work with the second one, if anyone could give me a nudge in the right direction, I would greatly appreciate it.

Answer 1

By C-S $$\left(\dfrac{1}{n}\sum_{k=1}^nx^2_k\right)^\dfrac{1}{2}=\frac{1}{n}\sqrt{\sum_{k=1}^n1^2\sum_{k=1}^nx_k^2}\geq\frac{1}{n}\sqrt{\left(\sum_{k=1}^nx_k\right)^2}=\left|\frac{1}{n}\sum_{k=1}^nx_k\right|.$$

Answer 2

Let ${\bf x}=(x_1,\ldots,x_n)$ and ${\bf y}=(1/n,\ldots,1/n)$. Use Cauchy-Schwartz inequality, \begin{align*} \lVert{\bf x}\rVert\lVert{\bf y}\rVert&=\left(\sum_{k=1}^{n}{x_k^2}\right)^{1/2}\left(n\cdot\frac{1}{n^2}\right)^{1/2}=\left(\frac{1}{n}\sum_{k=1}^{n}{x_k^2}\right)^{1/2} \\ &\geq|\langle{\bf x},{\bf y}\rangle|=\left\lvert\frac{1}{n}\sum_{i=1}^{n}{x_k}\right\rvert. \end{align*} This answer is essentially the same as Michael's, except using different notations in the intermediate steps.