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? :)
 A: 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
$$a+b+c+\frac{a+b+c}{3}\ge 4\left(\frac{abc(a+b+c)}{3}\right)^{1/4}\iff \frac{a+b+c}{3}\ge\sqrt[3]{abc}.$$
A: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = \tfrac12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \ge 0$$
A: Using convexity of $\exp:$
$$
\frac 13(a+b+c)
= \frac 13(\exp \log a+\exp \log b+\exp \log c)
\\\ge \exp \frac 13( \log a+ \log b+ \log c)
= (abc)^{1/3}
$$

Bonus: it does not only works for $n=3$ but for any $n$.
A: since $\ln(x)$ is a concave function on the positive reals, we have, by Jensen's inequality:
$$
E(\ln(X)) \le \ln(E(X))
$$
where $X$ is a random variable taking the values $a, b $ or $c$ with equal probability
A: powering by three we obtain
${a}^{3}+3\,{a}^{2}b+3\,{a}^{2}c+3\,a{b}^{2}-21\,abc+3\,a{c}^{2}+{b}^{3
}+3\,{b}^{2}c+3\,b{c}^{2}+{c}^{3}
\geq 0$
with $b=a+u,c=a+v$ we get
$\left( 9\,{u}^{2}-9\,uv+9\,{v}^{2} \right) a+{u}^{3}+3\,{u}^{2}v+3\,u
{v}^{2}+{v}^{3}
\geq 0$ which is true.
A: If we can use other inequalities, by Rearrangement 
$$a.a.a+b.b.b+c.c.c \ge a.b.c+b.c.a+c.a.b$$
generalises.
