Prove that $||v||^2 ≥ \dfrac{1}{n}$ - alternative way

Question

Let $v$ = $ \begin{align} & \begin{bmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{bmatrix} \end{align} ∈ \Bbb R^n$ s.t $x_1, ...,x_n ≥ 0$ and $\sum_{i=1}^{n}x_i =1$. Prove that $||v||^2 ≥ \dfrac{1}{n}$. The official solution uses Cauchy-Schwartz inequality but I solved it in a much simpler way. By definition, The standard norm is $||v||^2$ = $(\sqrt{\langle v,v \rangle})^2= $$(\sqrt{(\sum_{i=1}^{n}(x_i)^2})^2$ =$\sum_{i=1}^{n}x_i$ $*$ $\sum_{i=1}^{n}x_i$ (finite sum) = $1$ ≥ $\dfrac{1}{n}$ for every $n≥1$. There is something wrong here? It looks too naive.

Answer 1

-Take into account that I am just a student so I hope that my answer is correct.

-You have a more simple prove.

1- We know that $\sqrt{\frac{\sum_{i=1}^n x_i^2}{n}} \geq \frac{\sum_{i=1}^n x_i}{n} = \frac{1}{n}$ as it is given $\sum_{i=1}^n x_i = 1 $.
-This cause that $\sqrt{\frac{\sum_{i=1}^n x_i^2}{n}} \geq \frac{\sum_{i=1}^n x_i}{n} = \frac{1}{n} \Rightarrow \sqrt{\sum_{i=1}^n x_i^2} \geq \frac{1}{\sqrt{n}} $

2- Now as $x^2$ is an increasing function fofr $x \geq 0$ we have that $0 \leq x_1 \leq x_2 \Rightarrow x_1^2 \leq x_2^2$. Apply to here it gives us $\| v \|^2 = \sum_{i=1}^n x_i^2 =(\sqrt{\sum_{i=1}^n x_i^2})^2 \geq (\frac{1}{\sqrt{n}})^2= \frac{1}{n}$

Q.E.D.