Classical Inequalities addicts can do this. Part 2

Let's generalize again!

let $a_1 + a_2 + ... + a_n = x \\ (a_1)^2 + (a_2)^2 + ... + (a_n)^2 = y$

our aim is to minimize $f(a_1 , a_2 , ... , a_n) = (a_1 + a_2 + a_3 + ... + a_{n-1})^2 + (a_2 + a_3 + a_4 + ... + a_n)^2 + (a_3 + a_4 + ... + a_n + a_1)^2 + ... + (a_n + a_1 + a_2 + ... + a_{n-2})^2$ .

This can be simplify to: $f(a_1 , a_2 , ... , a_n) = (x - a_n)^2 + (x - a_1)^2 + (x - a_2)^2 + ... + (x - a_{n-1})^2 \\ = nx^2 - 2x^2 + y \\ = (n-2)x^2 + y$

By cauchy, $\rightarrow x^2 \le n[(a_1)^2 + (a_2)^2 + ... + (a_n)^2] \rightarrow x^2 \le ny$

By combining the trivial inequality and $x^2 \le ny$ ,

$\rightarrow 0 \le x^2 \le ny \\ 0 \le (n-2)x^2 \le n(n-2)y \\ y \le f(a_1 , a_2 , ... , a_n) \le \left[ (n-1)^2 \right] y$ .

Then,

$\rightarrow min(f(a_1 , a_2 , ... , a_n)) = y \\ max(f(a_1 , a_2 , ... , a_n)) = \left[ (n-1)^2 \right] y$ .

now, letting $n = 5 \\ a_1 = 5x \\ a_2 = 4y \\ a_3 = 3z \\ a_4 = 2a \\ a_5 = b \\ y = (5x)^2 + (4y)^2 + (3z)^2 + (2a)^2 + b^2 = \cfrac{100}{3}$

$\rightarrow f(5x,4y,3z,2a,b) = (5x+4y+3z+2a)^2 + (4y+3z+2a+b)^2 + (3z+2a+b+5x)^2 + (2a+b+5x+4y)^2 + (b+5x+4y+3z)^2 \\ min(f(5x,4y,3z,2a,b)) = y = \cfrac{100}{3} \\ max(f(5x,4y,3z,2a,b)) = [(n-1)^2]y = [(5-1)^2] \left[ \cfrac{100}{3} \right] = \cfrac{1600}{3}$

$\therefore max(f(5x,4y,3z,2a,b)) - min(f(5x,4y,3z,2a,b)) = \cfrac{1600}{3} - \cfrac{100}{3} = \boxed{500}$

A solution using matrices.

We are given that $\sum_{i=1}^5 x_i^2=r^2$ , where $x_1=5x,\ x_2=4y,\ x_3=3z,\ x_4=2a,x_5=b,\ r^2=\frac{100}{3}$ . Let, $\mathbf{x}=[x_1\ x_2\ \cdots\ x_n]^T$ , where $n=5$ . The function $f$ can be then expressed as $\sum_i^n(x-x_i)^2$ , where $x=\sum_{i=1}^n x_i$ . This can be further expressed as $f=\mathbf{x}^T\left(\sum_{i=1}^n (\mathbf{1}_i-\mathbf{e}_i)(\mathbf{1}_i-\mathbf{e}_i)\right)\mathbf{x}$ where $\mathbf{1}=[1 \ 1\ 1\cdots\ 1]^T\in \mathbb{R}^n$ , and $\mathbf{e}_i=[0\ 0\ \cdots\ 0\ 1\ 0\ \cdots\ 0]^T$ , with the $0$ at the $i$ th coordinate. Clearly, $\sum_{i=1}^n (\mathbf{1}_i-\mathbf{e}_i)(\mathbf{1}_i-\mathbf{e}_i)=n\mathbf{1}\mathbf{1}^T-\mathbf{1}\sum_{i=1}^n \mathbf{e}_i^T-\sum_{i=1}^n\mathbf{e}_i\sum_{i=1}^n \mathbf{1}^T+\sum_{i=1}^n\mathbf{e}_i\mathbf{e}_i^T=(n-2)\mathbf{1}\mathbf{1}^T+\mathbf{I}$ where $\mathbf{I}$ is the identity matrix. Thus, $f(\mathbf{x})=(n-2)(\mathbf{x}^T\mathbf{1})^2+\mathbf{x}^T\mathbf{x}=(n-2)\left(\sum_{i=1}^n x_i\right)^2+r^2$ Now, it can be seen that the minimum value of $(\sum_{i=1}^n x_i)^2$ is $0$ as it can be achieved as follows: If $n$ is even, then by choosing $n/2$ entries each equal to $+r/{\sqrt{2n}}$ , and $n/2$ entries $-r/{\sqrt{2n}}$ , and if $n$ is odd, then by choosing $(n-1)/2$ entries, $r/\sqrt{2n}$ , $(n-1)/2$ entries $-r/\sqrt{2n}$ and one entry $0$ . The maximum value of $(\sum_{i=1}^n x_i)^2$ is obtained from Cauchy-Scwartz to be $nr^2$ , which is achieved when all the $n$ entries are equal to $r/\sqrt{n}$ . Thus, $\max f-\min f=(n-2)nr^2-0=n(n-2)r^2$ , which, for the given question, evaluates to $\boxed{500}$ .

Define the matrix $A= \begin{pmatrix} 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 &1 \\ 1 & 0 & 1 & 1 &1 \\ 1 & 1 & 0 & 1 & 1\\ 1 & 1 & 1 &0 &1 \end{pmatrix}$ and the vector $\vec{r} = \begin{pmatrix} 5x \\ 4y \\ 3z \\ 2a \\ b \end{pmatrix}.$ Then we can write $f(\vec{r})= ||A\vec{r}||^2= \vec{r}'A'A\vec{r}$ with the constraint, $\vec{r}'\vec{r}= \frac{100}{3}.$ Using the Courant-Fischer theorem , we immediately have for non-zero $\vec{r}$ $\max f(\vec{r}) = \lambda_{\max}(A'A) \times \frac{100}{3}= 16 \times \frac{100}{3}$ and, $\min f(\vec{r})= \lambda_{\min}(A'A) \times \frac{100}{3}= 1\times \frac{100}{3},$ where $\lambda_{\max}(A'A)$ and \lambda_\min(A'A) denote the largest and smallest eigenvalue of the real symmetric matrix $A'A$ . Hence, the answer follows.