Finding points that satisfy $f(a) = \sup f(x)$ Choose positive real numbers $\alpha_1,\ldots,\alpha_n$, $n$ such that $\sum_{i=1}^n \alpha_i = 1$ and let $$f: [0,\infty)^n \to \mathbb R$$
$$x=(x_1, \ldots, x_n) \mapsto x_1^{\alpha_1} \cdots x_n^{\alpha_n}.$$
Define also $$C= \{x=(x_1, \ldots, x_n) \in [0,\infty)^n\text{ such that }\sum_{i=1}^n \alpha_ix_i=1 \}.$$ 
a) Show that there exists $a=(a_1 ,\ldots, a_n)$ such that $f(a) = \sup_{x\in C} f(x)$, and $\alpha_i>0$ for any $i$.
b) Find all points such that $f(a) = \sup_{x \in C} f(x)$.  Deduce that for any $x \in C$, $x_1^{\alpha_1} \cdots x_n^{\alpha_n} \le 1.$
c) Deduce that for any $x\in[0,\infty)^n$, $x_1^{\alpha_1}\cdots x_n^{\alpha_n} \le \sum_{i=1}^n {\alpha_ix}_i$.  When does equality hold?
Edit: Hmm...since the $\alpha_i$'s are positive real numbers and sum to 1, then we know that each $\alpha_i$ is strictly < 1 , that is, 0< $\alpha_i$<1 for 1$\le i \le n$.
Any help to get me started would be greatly appreciated.
Thanks in advance,
 A: $f(x_1,x_2,\cdots,x_n) = x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}$ subject to:
$\displaystyle \sum_{i=1}^n x_i\alpha_i = 1$. Use Lagrange Multiplier:
$\dfrac{\partial f}{\partial x_i} = \lambda\alpha_i \Rightarrow x_1^{\alpha_1}x_2^{\alpha_2}\cdots \alpha_ix_i^{\alpha_i-1}\cdots x_n^{\alpha_n} = \lambda\alpha_i,\forall i = \overline{1,n} \Rightarrow x_i = \dfrac{1}{\lambda}, \forall i = \overline{1,n} \Rightarrow x_1 = x_2 =\cdots = x_n = \dfrac{1}{\alpha_1+\alpha_2+\cdots + \alpha_n} = 1$. Thus $a = (1,1,\cdots,1)$, and $f(a) = \text{sup}\{f(x): x \in C\}=1 \Rightarrow x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n} \leq f(a) = 1$.
For part $c)$ we have: $\left(\dfrac{x_1}{\displaystyle \sum_{i=1}^n \alpha_ix_i}\right)^{\alpha_1}\cdot \left(\dfrac{x_2}{\displaystyle \sum_{i=1}^n \alpha_ix_i}\right)^{\alpha_2}\cdots \left(\dfrac{x_n}{\displaystyle \sum_{i=1}^n \alpha_nx_n}\right)^{\alpha_n} \leq 1 \Rightarrow \displaystyle \prod_{i=1}^n  x_i^{\alpha_i} \leq \left(\displaystyle \sum_{i=1}^n \alpha_ix_i\right)^{\displaystyle \sum_{i=1}^n \alpha_i} = \displaystyle \sum_{i=1}^n \alpha_ix_i$.
