Sum of the squares of twelve real numbers which add to $1.$ Let $a, b, c,\cdots, l$ be real numbers such that $a + b + c \dots + l = 1.$ Find the minimum value of $$a^2 + b^2 + c^2 + \dots + l^2.$$
My first glance at this problem, I would assume the minimum would be $1,$ gotten by a bunch of zeros, and then a $1.$ I'm not sure if this is correct, and is there a rigorous proof directed toward this?

Minimum occurs when $a_1 = a_2 = a_3 = \ldots = a_{12} = \frac{1}{12}$, and the minimum such value is $\frac{1}{12}$. To prove this, consider shifting the variables by setting $b_i = a_i - \frac{1}{12}$. We are then trying to minimise
$$\left(b_1 + \frac{1}{12}\right)^2 + \left(b_2 + \frac{1}{12}\right)^2 + \ldots + \left(b_{12} + \frac{1}{12}\right)^2$$
subject to $b_1 + b_2 + \ldots + b_{12} = 0$. We then have,
\begin{align*}
&\left(b_1 + \frac{1}{12}\right)^2 + \left(b_2 + \frac{1}{12}\right)^2 + \ldots + \left(b_{12} + \frac{1}{12}\right)^2 \\
= \; &b_1^2 + b_2^2 + \ldots + b_{12}^2 + \frac{1}{6}(b_1 + b_2 + \ldots + b_{12}) + \frac{1}{12} \\
= \; &b_1^2 + b_2^2 + \ldots + b_{12}^2 + \frac{1}{12}.
\end{align*}
Clearly the above function has a minimum of $\frac{1}{12}$, and is achieved if and only if $b_1 = b_2 = \ldots = b_{12} = 0$ (which satisfies the constraint), i.e. when $a_1 = a_2 = \ldots = a_{12} = \frac{1}{12}$.
