You can use the Cauchy-Schwarz inequality to prove that
$$\frac{x + y}{2} \geqslant \sqrt{xy} \tag{$\dagger$}$$
as you have done, if you replace $x^2$ and $y^2$ with $x$ and $y$.
We shall extend the AM–GM inequality from two to four variables, then reduce it to three. Applying $(\dagger)$ twice on the arithmetic mean of four variables, $(w + x + y + z)/4$, gives
$$\frac{(w + x) + (y + z)}{4} \geqslant \frac{\sqrt{wx} + \sqrt{yz}}{2}\geqslant \sqrt[4]{wxyz}.$$
Now let $w = \frac{1}{3}(x + y + z)$ so that
$$\begin{align*}
(x + y + z)/3 &\geqslant \sqrt[4]{(x + y + z)/3 \cdot xyz}\\
\frac{1}{3^{3/4}}(x + y + z)^{3/4} &\geqslant \sqrt[4]{xyz}\\
\frac{x + y + z}{3} &\geqslant \sqrt[3]{xyz},\\
\end{align*}$$
as desired.
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}
$.
Best Answer
Because of you've done with four variables, you can use it like this.
$$x+y+z + \frac{x+y+z}{3} \geq 4\sqrt[4]{xyz(\frac{x+y+z}{3})}$$
then $$\frac{4}{3}(x+y+z) \geq 4\sqrt[4]{xyz(\frac{x+y+z}{3})}$$
or, $$\frac{x+y+z}{3} \geq \sqrt[4]{xyz(\frac{x+y+z}{3})},$$
$$\left(\frac{x+y+z}{3}\right)^3 \geq xyz.$$