Proving AM-GM for the special case $n=3$ I know the AM-GM inequality and its proof which is relatively complex, though the case for $n=2$ is quite simple. However, I don't know of any special easier proof for the case $n=3$, specifically:
$$\frac{a+b+c}3\ge \sqrt[3]{abc}$$
What is the most elegant proof for this? :)

The case for $n=3$ can be proved by using the cases for $n=2,4$.
For $p,q\gt 0$, we have$$(\sqrt p-\sqrt q)^2\ge0\iff \frac{p+q}{2}\ge\sqrt{pq}.$$
So, we have for $s,t,u,v\gt 0,$$$s+t\ge 2\sqrt{st},\ \ \ u+v\ge 2\sqrt{uv}.$$
Hence, we have
$$s+t+u+v\ge 2\sqrt{st}+2\sqrt{uv}\ge 2\sqrt{2\sqrt{st}\cdot 2\sqrt{uv}}=4(stuv)^{1/4}.$$
Here, setting $s=a,t=b,u=c,v=\frac{a+b+c}{3}$ gives us
$$a+b+c+\frac{a+b+c}{3}\ge 4\left(\frac{abc(a+b+c)}{3}\right)^{1/4}\iff \frac{a+b+c}{3}\ge\sqrt[3]{abc}.$$
