Sum of the squares of twelve real numbers which add to $1.$

Question

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?

Answer 1

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 $a_1 = a_2 = \ldots = a_{12} = \frac{1}{12}$.

Answer 2

$a+b+c+\cdots+l = 1$ is a plane, having $a$-intercept, $b$-intercept, ..., $l$-intercept $1$. The closest point to the origin is in the $(1,1,1,\cdots,1)$ direction. The value $a^2 + b^2 + c^2 + \cdots+ l^2$ is the square of the Euclidean distance of the point $(a,b,c,d,\dots, l)$ from the origin.

As observed, the nearest point on the plane has all coordinates equal and summing to $1$, i.e., $(1/12, 1/12, \cdots, 1/12)$. The sum of the squares of those coordinates is $1/12$.

Answer 3

From AM-QM inequality, we have $a^2+b^2+\cdots+l^2 \geq \frac{1}{12}(a+b+\cdots+l)^2 = \frac{1}{12}$, where the equality holds if and only if $a = b=\cdots =l = \frac{1}{12}$.