1. ## Minimize

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

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

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

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

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

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

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

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

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

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

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

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

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