summation question

Dec 2009
21
0
i want to show

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

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

any idears

thankss
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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.)
 
Nov 2009
485
184
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 \(\displaystyle g(a_1,\ldots, a_n)=a_1+\ldots+a_n=1\).
 
Dec 2009
21
0
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
 

awkward

MHF Hall of Honor
Mar 2008
934
409
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
\(\displaystyle a_1 + a_2 + \dots + a_n \leq |a_1| +| a_2| + \dots + |a_n| < 1/n + 1/n + \dots +1/n = 1\)