If $a_i>o$ then $(a_1a_2\cdots a_{2^n})^{1/2^n}\leq \frac{a_1+a_2+\cdots+a_{2^n}}{2^n}$ I need help to prove this inequality, I have no idea how to proceed with the inductive step:

$$a_1,a_2,\ldots,a_{2^n}>0 \Longrightarrow(a_1a_2\cdots a_{2^n})^{1/2^n}\leq \frac{a_1+a_2+\cdots+a_{2^n}}{2^n},\forall n\in\mathbb{N}$$

Any help would be appreciated.
 A: Once you get the first two steps, it is straightforward.
Note that
$$a_1a_2 \leq \left[\frac{a_1+a_2}{2}\right]^2$$
and
$$(a_1a_2)(a_3a_4)\leq\left[\frac{a_1+a_2}{2}\right]^2\left[\frac{a_3+a_4}{2}\right]^2 = \left[\left(\frac{a_1+a_2}{2}\right)\left(\frac{a_3+a_4}{2}\right)\right]^2\\ \leq \left[\frac1{2}\left[\left(\frac{a_1+a_2}{2}\right)+\left(\frac{a_3+a_4}{2}\right)\right]\right]^4 =\left[\frac1{4}\sum_{i=1}^{4}a_i\right]^4.$$
Continue with induction to get
$$\left[\prod_{i=1}^{2^n}a_i\right]^{1/2^n} \leq \frac1{2^n}\sum_{i=1}^{2^n}a_i.$$
A: The trick is to take advantage of the fact that the number of elements is a power of $2$ by splitting the factors into two equal parts and using the induction hypothesis on both halves. We want to show that 
$$(a_1a_2\cdots a_{2^{n+1}})^\frac{1}{2^{n+1}}\leq \frac{a_1+a_2+\ldots+a_{2^{n+1}}}{2^{n+1}}.$$
On the other hand, by the induction hypothesis we know that
$$(a_1a_2\cdots a_{2^n})^\frac{1}{2^n}\leq \frac{a_1+a_2+\ldots+a_{2^n}}{2^n}$$
and similarly
$$(a_{n+1}a_{n+2}\cdots a_{2^{n+1}})^\frac{1}{2^n}\leq \frac{a_{n+1}+a_{n+2}+\ldots+a_{2^{n+1}}}{2^n}.$$
What happens when you multiply these last two inequalities?
