# summation question

• Jun 20th 2010, 09:35 AM
silk23
summation question
i want to show

1/(n^2) <= (a1^2) + (a2^2) + (a3^2) + ..... + (an^2)

given that a1 + a2 + .... + an =1

any idears

thankss
• Jun 20th 2010, 09:40 AM
undefined
Quote:

Originally Posted by silk23
i want to show

1/(n^2) <= (a1^2) + (a2^2) + (a3^2) + ..... + (an^2)

given that a1 + a2 + .... + an =1

any idears

thankss

Edit: Sorry I misread the problem. I'll think about it. (But someone else will probably answer before I come up with anything.)
• Jun 20th 2010, 09:53 AM
roninpro
It suffices to show $\displaystyle \frac{1}{n}\leq a_1^2+\ldots+a_n^2$ since $\displaystyle \frac{1}{n^2}\leq \frac{1}{n}$.

How about using calculus? Lagrange Multipliers might do the trick. Optimize $\displaystyle f(a_1,\ldots, a_n)=a_1^2+\ldots+a_2^2$ subject to $\displaystyle g(a_1,\ldots, a_n)=a_1+\ldots+a_n=1$.
• Jun 20th 2010, 10:30 AM
silk23
ok so let F = (a1^2) + (a2^2) + (a3^2) + ..... + (an^2) + y( a1 + ... + an -1)

df/dai = 2ai +y = 0

ai= -y/2

df/dfy = a1 + ... + an -1 =0

so y= -2/n so

ai = 1/n maxamises the (a1^2) + (a2^2) + (a3^2) + ..... + (an^2) so maximum is n*(1/n^2) =1/n

and 1/(n^2) < 1/n

thanks
• Jun 21st 2010, 04:57 PM
awkward
Here is an alternative, non-calculus approach. Assume the contrary, i.e.

$\displaystyle a_1^2 + a_2^2 + \dots + a_n^2 < 1/n^2$.

Then for any i,
$\displaystyle a_i^2 < 1/n^2$

so
$\displaystyle |a_i| < 1/n$
and
$\displaystyle a_i \leq |a_i|$
so
$\displaystyle a_1 + a_2 + \dots + a_n \leq |a_1| +| a_2| + \dots + |a_n| < 1/n + 1/n + \dots +1/n = 1$