For $1 \leq k\leq n$ if $x_k$ is a non-zero real number and $x_1^2+\ldots +x_n^2=X.$ Is it true $x_1+\ldots +x_n \leq X^{\frac{1}{2}} ?$

Question

Question: For $1 \leq k\leq n$ if $x_k$ is a non-zero real number and $x_1^2+\ldots +x_n^2=X.$ Is it true $x_1+\ldots +x_n \leq X^{\frac{1}{2}} ?$

I would like to think this is straightforward from the Cauchy-Schwarz inequality $^1$ - but I am not certain. I think I can write $\left(x_1 +\ldots +x_n\right)^2 \leq x_1^2+\ldots +x_n^2\leq X.$ I can then take square roots on the LHS and RHS in order to get $x_1+\ldots +x_n \leq X^{\frac{1}{2}}.$

Note $x_k$ can be negative.

Answer 1

By Cauchy-Schwarz inequality, $$\left(x_1+x_2+\cdots+x_n\right)^2 \le \color{red}{n}(x_1^2+x_2^2+\cdots x_n^2) = n X.$$ Therefore, $$x_1 + x_2 + \cdots + x_n \le |x_1 + x_2 + \cdots + x_n| \le \sqrt{nX}.$$