# Find the largest and smallest

• Nov 9th 2007, 01:03 AM
perash
Find the largest and smallest
The sum of the non-negative real numbers $s_1, s_2,..., s_{2004}$ is 2 and
$s_1s_2+s_2s_3+ ...+ s_{2003} s_{2004}+ s_{2004}s_1=1$.
Find the largest and smallest possible values of
$S=s_1^2+s_2^2+...+s_{2004}^2$
• Nov 9th 2007, 03:10 AM
CaptainBlack
Quote:

Originally Posted by perash
The sum of the non-negative real numbers $s_1, s_2,..., s_{2004}$ is 2 and
$s_1s_2+s_2s_3+ ...+ s_{2003} s_{2004}+ s_{2004}s_1=1$.
Find the largest and smallest possible values of
$S=s_1^2+s_2^2+...+s_{2004}^2$

$\sum_{i=1}^{2400} s_i =2$

Now square:

$\left(\sum_{i=1}^{2400} s_i\right)^2 = \sum_{i=1}^{2400} s_i^2 +2\sum_{i \ne j} s_i s_j = \sum_{i=1}^{2400} s_i^2 +2 =4$

So take it from there.

RonL
• Dec 18th 2007, 07:29 PM
JaneBennet
Quote:

Originally Posted by CaptainBlack
$\left(\sum_{i=1}^{2400} s_i\right)^2 = \sum_{i=1}^{2400} s_i^2 +\color{red}2\sum_{i \ne j} s_i s_j \color{black}= \sum_{i=1}^{2400} s_i^2 +\color{red}2\color{black} =4$

No. $\sum_{i \ne j}{s_i s_j}$ is not the same as $
s_1s_2+s_2s_3+\ldots+ s_{2003} s_{2004}+ s_{2004}s_1
$
. The former contains terms like $s_2s_4$ that are not present in the latter. Rather,

$S\ =\ s_1^2+s_2^2+\ldots+s_{2004}^2$

$=\ (s_1+\ldots+s_{2004})^2-2\sum_{i \ne j}{s_i s_j}$

$\leq\ (s_1+\ldots+s_{2004})^2-2(s_1s_2+s_2s_3+\ldots+ s_{2003} s_{2004}+ s_{2004}s_1)$

$=\ 2^2-2\ =\ 2$

For the lower bound, the Cauchy–Schwarz inequality gives

$|s_1s_2+s_2s_3+\ldots+ s_{2003} s_{2004}+ s_{2004}s_1|\ \leq\ (s_1^2+s_2^2+\ldots+s_{2004}^2)^{\frac{1}{2}}(s_2^ 2+s_3^2+\ldots+s_{2004}^2+s_1^2)^{\frac{1}{2}}
$

so $S\ \geq\ 1$.