How to prove the sum of squares is minimum? Given $n$ nonnegative values. Their sum is $k$.
$$
x_1 + x_2 + \cdots + x_n = k
$$
The sum of their squares is defined as:
$$
x_1^2 + x_2^2 + \cdots + x_n^2
$$
I think that the sum of squares is minimum when $x_1 = x_2 = \cdots = x_n$. But I can't figure out how to prove it. Can anybody help me on this? Thanks.

@Quinn Culver
Quinn Culver's answer is not enough.
It's not enough to say that $x_i=k/n$ is the local optima. Based on the Second Order Sufficiency Conditions, we need to prove $y^T\nabla_{xx}J y>0$ for all $y\ne0$ with $\nabla h(x)^Ty=0$. After some computation, we find $\nabla_{xx}J=2I$, where $I$ is the identity matrix and the vector $y$ satisfies $\nabla h(x)^Ty=0$ is $y^T1=0, \ y\ne 0$. In the end, we find $y^T\nabla_{xx}J y=2y^Ty >0$. Therefore, we prove that $x_i=k/n$ is the local optima. Furthermore, because $x_i=k/n$ is the unique solution, therefore, it is the global optima.  
