A question of summation.

Question

Let $X_1,X_2.....X_n$ be positive numbers such that $X_1+X_2.....+X_n=17$. Find the minimum of $X_1^2+X_2^2.....+X_n^2$. (Obviously in terms of n) ^_^

Answer 1

Using the $AM$-$QM$ (Arithmetic Mean - Quadratic Mean) inequality:

$$ \begin{align} \frac{\sum^{n}_{i=1}X_i}{n} &\leq \sqrt{\frac{\sum^{n}_{i=1}{X_i}^2}{n}}\\ \frac{17}{n} &\leq \sqrt{\frac{\sum^{n}_{i=1}{X_i}^2}{n}}\\ \frac{289}{n^2} &\leq \frac{\sum^{n}_{i=1}{X_i}^2}{n}\\ \frac{289}{n} &\leq \sum^{n}_{i=1}{X_i}^2\\ \end{align} $$

Equality holds when $X_1 = X_2 = \cdots = X_{n} = \frac{17}{n}$

EDIT:

The quadratic mean of a set of numbers $x_1,x_2,\cdots ,x_n$ is $\sqrt{\frac{\sum^{n}_{i=1}{x_i}^2}{n}}$.

Given that the question is asking for a minimum value, it is likely that an inequality is involved.

If another method is to be used, another idea that comes off the top of my head is using the following identity:

$$ \sum^{n}_{i=1}X_i^2 = (\sum^{n}_{i=1}X_i)^2 - \sum^{n}_{i=1}\sum^{n}_{j=i+1}2X_iX_j $$

To minimise the LHS, $\sum^{n}_{i=1}\sum^{n}_{j=i+1}2X_iX_j$ must be maximised as $(\sum^{n}_{i=1}X_i)^2$ is constant.

By the $AM$-$GM$ inequality (this is more elementary):

$$ 2x_ix_j \leq x_i^2 + x_j^2 $$

Thus:

$$ \begin{align} \sum^{n}_{i=1}X_i^2 &= (\sum^{n}_{i=1}X_i)^2 - \sum^{n}_{i=1}\sum^{n}_{j=i+1}2X_iX_j\\ &\geq (17)^2 - \sum^{n}_{i=1}(n-1) \cdot X_i^2\\ n \cdot \sum^{n}_{i=1}X_i^2 &\geq 289 \\ \sum^{n}_{i=1}X_i^2 &\geq \frac{289}{n} \end{align} $$

Answer 2

$\sqrt{(X_1^2+...+X_n^2)/n)}$ $\ge$ $(X_1+...+X_n)/n=17/n$

min of $X_1^2+...+X_n^2$ = $289/n$

Answer 3

The answer may also be found by geometric considerations.

The "summation" equation:

$$ X_1 + X_2 + \ldots + X_n = 17 $$

describes a hyperplane in $n$-dimensional space (a line in $2$-dimensional space when $n=2$). We are asked to find the radius $r$ of a hypersphere centered at the origin which is tangent to this hyperplane, for the square of that radius is the minimum of $X_1^2 + X_2^2 + \ldots + X_n^2$ of any point in the hyperplane.

Draw a hyperplane parallel to the given one above, but passing through the origin:

$$ X_1 + X_2 + \ldots + X_n = 0 $$

The point of tangency we seek lies on a line perpendicular to both hyperplanes and passing through the origin. The distance between the two hyperplanes is the radius of the hypersphere we described, and one such radius is the line segment between the origin and the point of tangency.

Note that the equation of the hyperplane through the origin says that:

$$ (1,1,\ldots,1) \cdot (X_1,X_2,\ldots,X_n) = 0 $$

for any point $(X_1,X_2,\ldots,X_n)$ lying in this hyperplane. It follows that the perpendicular line consists of just multiples of $(1,1,\ldots,1)$.

To find the point of tangency, and thus the square of the corresponding radius, we need only locate the multiple of $(1,1,\ldots,1)$ which lies in the original hyperplane. That is, for what constant $c$ does:

$$ c + c + \ldots + c \equiv nc = 17 $$

Obviously $c = 17/n$. Thus the square of the radius (the squared length of the line segment from $(c,c,\ldots,c)$ to the origin) is $n(17/n)^2 = 289/n$, in agreement with the other answers.