I am self-studying and currently reading How to Prove it by Velleman. I tried to prove the above by induction (I proved that this holds true for $n=2$), but I think my proof is wrong. I only started reading on proofs and induction yesterday and while it makes sense on paper,it still gives me a hard time to use. So here is my proof:
$$ \frac {a_1+a_2+...+a_{2^{n+1}}}{2^{n+1}}={\frac{1}{2}\frac{a_1+a_2+...a_{2^{n}}}{2^{n}}}+\frac{a_{2^n+1}+a_{2^n+2}+...+a_{2^{n+1}}}{{2^{n+1}}} \quad \frac {a_1+a_2+...+a_{2^{n+1}}}{2^{n+1}} \geq \frac{1}{2} \sqrt[2^n]{a_1a_2...a_{2^n}}+\sqrt[2^{n+1}]{a_1a_2...a_{2^{n+1}}}-\frac{1}{2}\sqrt[2^n]{a_1a_2...a_{2^n}} \quad \frac {a_1+a_2+...+a_{2^{n+1}}}{2^{n+1}} \geq \sqrt[2^{n+1}]{a_1a_2...a_{2^{n+1}}} $$
Also sorry for any formatting mistakes, this is my first time trying Latex. This is also why the proof skips a few steps at the beginning.
There are only $2^n$ numbers from $2^n+1$ to $2^{n+1}$. So the last part on the first line should be $$\frac12\frac{a_{2^n+1}+a_{2^n+2}+\cdots+a_{2^{n+1}}}{2^n}$$ very similar to the first part.
Then, after applying the AM-GM on each of the two parts, you will need to apply the AM-GM again to combine the two parts into one.