Short "beautiful" proof for simple fact wanted

Question

I'd like to find a cute proof for the following fact:

Let $x_1, \dotsc, x_n \in \mathbb{N}$ be such that $\sum_{i=1}^n x_i = X$ for some fixed $X \in \mathbb{N}$ and $x_i \leq v$ for all $1 \leq i \leq n$. Then $$\sum_{i=1}^n x_i^2 \leq \frac{X}{v} v^2=Xv$$

Thanks for any input.

Answer 1

$$Xv=\sum_{i=1}^nx_iv\geq\sum_{i=1}^nx_ix_i=\sum_{i=1}^nx_i^2$$

Answer 2

Proceed as when maximizing entropy: Use lagrange multipliers to maximize $\sum_i x_i^2$ subject to the constraint $\sum_{i=1}^n x_i = X$. You then get a maximum for $x_i=\frac{X}{n}$.

This means:

$$\sum_{i=1}^n x_i^2 \leq \sum_{i=1}^n\frac{X^2}{n^2}=\frac{X^2}{n}=X\frac{X}{n}\leq X v$$

Where the last inequality comes from the fact that $x_i=\frac{X}{n}$ satisfies the requirements.