if $\sum\limits_{i=1}^n{x_i} = 1$, how do you choose the $x_i$'s such that $\sum\limits_{i=1}^n{x_i^2}$ is minimized? if $\sum\limits_{i=1}^n{x_i} = 1$, how do you choose the $x_i$'s such that $\sum\limits_{i=1}^n{x_i^2}$ is minimized. 
I have an intuition that each $x_i = \frac{1}{n}$, but I don't know how to prove it
 A: $$0\leq \sum_{k=1}^n\left(x_k-\frac1n\right)^2=\left(\sum_{k=1}^nx_k^2\right)-\frac1n$$ with equality if and only if $x_k=\tfrac1n$ for $k\in\{1,\ldots,n\}$.
A: Using the Cauchy-Schwarz inequality
$$1 = \left|\sum_{i = 1}^n x_i\right|^2 \leqslant \sum_{i = 1}^n x_i^2\sum_{i = 1}^n (1)^2 = n\sum_{i = 1}^n x_i^2.$$
Hence,
$$\sum_{i = 1}^n x_i^2 \geqslant \frac1{n},$$
and the minimum is attained when $x_i = 1/n$.
A: Your intution is correct, and you can prove it using a Lagrange multiplier: Setting the derivative of $\sum_ix_i^2-\lambda(\sum_ix_i-1)$ with respect to $x_j$ to zero yields $2x_j=\lambda$, and then your result follows from the normalisation condition.
A: Denote the origin by $O$, and let $H$ be the hyperplane given by $\sum_{i=1}^n x_i = 0$.  This question is asking for the point on $H$ closest to $O$.
The line $L$ given by $x_1=x_2=\cdots  = x_n$ passes through $O$ and is perpendicular to $H$.  By the Pythagorean Theorem, the shortest distance from $O$ to $H$ lies along the perpendicular.
Since the intersection of $L$ with $H$ is $x_i = 1/n$, this is the closest point.
A: $$1=1^2=\left(\sum_ix_i\right)^2=\sum_ix_i^2+\sum_{i<j}2x_ix_j\le\sum_ix_i^2+\sum_{i<j}(x_i^2+x_j^2)=n\sum_ix_i^2$$
where the inequality comes from the fact that $(x-y)^2=x^2-2xy+y^2\ge0$ (basically AM-GM inequality). We thus arrive at our result:
$$\sum_ix_i^2\ge\frac{1}{n}$$
