How is AM-GM supposed to be used to prove an inequality I have recently started learning proofs involving inequalities and came across the AM-GM inequality, which seems like a quite powerful tool. However, I am not sure I understand how to use this tool properly, and I was wondering if there are acceptable strategies to be aware of when making use of the AM-GM inequality.
I have also tried solving an inequality involving AM-GM to learn how to use this tool as I go, but I'm not sure if what I've done so far is a valid approach.

Here is the question:

Prove that if $x, y, z, w ≥ 0$,
 $\frac{x+y+z+w}{4} ≥ \sqrt[4]{xyzw}$ 

Here is what I have done so far:

I noticed that $\sqrt[4]{xyzw}$ = $\sqrt{\sqrt{xy}\sqrt{zw}}$, and then used AM-GM, like so:

I let $a = \sqrt{xy}$  $ b= \sqrt{zw}$
to make a use of $\frac{a+b}{2} ≥ \sqrt{ab}$. 

And, then, I subbed in the values:
$\frac{\sqrt{xy}+\sqrt{zw}}{2} ≥ \sqrt{\sqrt{xy}\sqrt{zw}}$. Now I have a feeling I'm supposed to somehow make a use of the inequality ($\frac{a+b}{2} ≥ \sqrt{ab}$) once more, however, I am not exactly sure how I should go about doing it.
 Any tips for a newbie like myself would be immensely helpful! (Perhaps a link I can refer to, an article, etc)
 Thanks a lot in advance for any help!
 A: Start with
$\dfrac{a+b}{2}
\ge \sqrt{ab}
$.
To prove this,
write it as
$\dfrac{a-2\sqrt{ab}+b}{2}
\ge 0
$,
and the left side is
$\dfrac{(\sqrt{a}-\sqrt{b})^2}{2}
\ge 0
$.
Then,
$\begin{array}\\
\dfrac{a+b+c+d}{4}
&=\dfrac{a+b}{4}+\dfrac{c+d}{4}\\
&=\dfrac{\dfrac{a+b}{2}}{2}+\dfrac{\dfrac{c+d}{2}}{2}\\
&\ge\dfrac{\sqrt{ab}}{2}+\dfrac{\sqrt{cd}}{2}\\
&=\dfrac{\sqrt{ab}+\sqrt{cd}}{2}\\
&\ge\sqrt{\sqrt{ab}\sqrt{cd}}\\
&=\sqrt{\sqrt{abcd}}\\
&=\sqrt[4]{abcd}\\
\end{array}
$
By induction on $n$,
with this technique
you can show that
$\dfrac{\sum_{k=1}^{2^n}a_k}{2^n}
\ge \sqrt[2^n]{\prod_{k=1}^n a_k}
$.
To show this is true
for any $m < 2^n$,
let
$a_j
=\dfrac{\sum_{k=1}^m a_k}{m}
$
for $j \gt m$
and see what happens.
As a matter of fact,
this was Cauchy's
original proof.
Here's the details (added later).
The left side is,
letting
$a = \dfrac{\sum_{k=1}^m a_k}{m}
$,
$\begin{array}\\
\dfrac{\sum_{k=1}^{2^n}a_k}{2^n}
&=\dfrac{\sum_{k=1}^{m}a_k}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a_k}{2^n}\\
&=\dfrac{\sum_{k=1}^{m}a_k}{m}\dfrac{m}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a}{2^n}\\
&=\dfrac{am}{2^n}+\dfrac{(2^n-m)a}{2^n}\\
&=\dfrac{am}{2^n}+\dfrac{2^na}{2^n}-\dfrac{ma}{2^n}\\
&= a\\
&=\dfrac{\sum_{j=1}^ma_j}{m}\\
\end{array}
$
Similarly,
the right side is,
letting
$a_j
=b
=\left(\prod_{k=1}^{m} a_k\right)^{1/m}
$
for $j > m$,
$\begin{array}\\
\sqrt[2^n]{\prod_{k=1}^{2^n} a_k}
&=\left(\prod_{k=1}^{2^n} a_k\right)^{1/2^n}\\
&=\left(\prod_{k=1}^{m} a_k\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\
&=\left(\prod_{k=1}^{m} a_k\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\
&=\left(b^m\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} b\right)^{1/2^n}\\
&=b^{m/2^n}\left(b^{2^n-m}\right)^{1/2^n}\\
&=b^{m/2^n}b^{(2^n-m)/2^n}\\
&=b\\
&=\left(\prod_{k=1}^{m} a_k\right)^{1/m}\\
\end{array}
$
Therefore
$a \ge b$
or
$\dfrac{\sum_{j=1}^ma_j}{m}
\ge \left(\prod_{k=1}^{m} a_k\right)^{1/m}
$.
A: $$\frac{x+y+z+w}{4}\ge\frac{\sqrt{xy}+\sqrt{zw}}{2}\ge\sqrt[4]{xyzw}$$
