i want to show

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

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

any idears

thankss

Printable View

- June 20th 2010, 10:35 AMsilk23summation 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 - June 20th 2010, 10:40 AMundefined
- June 20th 2010, 10:53 AMroninpro
It suffices to show since .

How about using calculus? Lagrange Multipliers might do the trick. Optimize subject to . - June 20th 2010, 11:30 AMsilk23
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 - June 21st 2010, 05:57 PMawkward
Here is an alternative, non-calculus approach. Assume the contrary, i.e.

.

Then for any i,

so

and

so