How does this prove AM-GM? Here is an extract from A less than B. The author claims to imply the AM-GM inequality with this reasoning, but I can't see how. So far the author has covered AM-GM, convexity, the smoothing principle and Jensen's inequality.
"Theorem 4: Let $f$ be a twice-differentiable function on an open interval $I$. Then $f$ is convex on $I$ if and only if $f''(x)\ge 0$ for all $ x \in I$.
For example, the AM-GM inequality can be proved by noting that $f(x)=\log(x)$ is concave; its first derivative is $1/x$ and its second $-1/x^2$. In fact, one immediately deduces a weighted AM-GM inequality..."
I understand the theorem, but not at all how it applies to AM-GM. Any enlightenment would be much appreciated!

Notice that $\ln{x}$ is defined only for $x>0$, which means that its second-order derivative is always negative, which means the function is concave. 
Let $f(x)=-\ln{x}$. We know that $f$ is convex. Thus, writing the Jensen's inequality for $f$ with the weights $t_{i}=\frac{1}{n}$ ($i=1,2,...,n$), we get:
$$-\ln\left(\frac{x_{1}+x_{2}+...+x_{n}}{n}\right)\leq\frac{1}{n}\sum^{n}_{i=1}(-\ln{x_{i}})=-\ln\sqrt[n]{x_{1}x_{2}\cdot...\cdot x_{n}}$$
now, since we know that $\ln$ is monotonic, we get:
$$\frac{x_{1}+x_{2}+...+x_{n}}{n}\geq\sqrt[n]{x_{1}x_{2}\cdot...\cdot x_{n}}.$$
