1. ## Minimize

Suppose you want to minimize $a_{1}^{2} + \cdots + a_{n}^{n}$ subject to the constraints $a_i >0$ and $a_1+ \cdots + a_n = 1$.

So in the Cauchy Schwarz Inequality do the following: Take $b_1 = \cdots = b_n = 1$. So we have:

$(a_{1}^{2}+ \cdots + a_{n}^{2})(b_{1}^{2} + \cdots b_{n}^{2}) \geq (a_{1}b_{1} + \cdots +a_{n}b_{n})^2$.

$a_{1}^{2} + \cdots + a_{n}^{2} \geq (a_1+ \cdots + a_n)^2$.

$a_{1}^{2} + \cdots + a_{n}^2 \geq 1$

So $a_1 = \cdots = a_n = \frac{1}{\sqrt{n}}$?

2. Originally Posted by Sampras
Suppose you want to minimize $a_{1}^{2} + \cdots + a_{n}^{n}$ subject to the constraints $a_i >0$ and $a_1+ \cdots + a_n = 1$.

So in the Cauchy Schwarz Inequality do the following: Take $b_1 = \cdots = b_n = 1$. So we have:

$(a_{1}^{2}+ \cdots + a_{n}^{2})(b_{1}^{2} + \cdots b_{n}^{2}) \geq (a_{1}b_{1} + \cdots +a_{n}b_{n})^2$.

$a_{1}^{2} + \cdots + a_{n}^{2} \geq (a_1+ \cdots + a_n)^2$.

$a_{1}^{2} + \cdots + a_{n}^2 \geq 1$

So $a_1 = \cdots = a_n = \frac{1}{\sqrt{n}}$?
Nevermind I forgot that $b_1^{2} + \cdots + b_{n}^{2} = n$.

3. Yes, you've got closer to the correct solusion, actually $a_1=a_2=\cdot \cdot \cdot =a_n=\frac{1}{n}$.
P.S. you are so humorous.