Proof of AM - GM inequality using concavity

The AM - GM (arithmetic - geometric mean ) inequality is expressed as follows:

$$\sum_{k=1}^{n}a_k \geq n \sqrt[n]{\prod_{k=1}^{n}a_k} \tag{1}$$

One proof was given by the French mathematician  Augustin Luis Cauchy in its lecture book that he had prepared for his students. The proof is based on induction , the so called "back and forth" form of induction. Since then many proofs of this inequality have been discovered. In this topic we give a proof based on the concavity of the $\log $ function.

Proof



The function $f(x) = \log x$ is concave on $(0, +\infty)$. Jensen's inequality states that if a function is concave then:

$$f\left ( \frac{a+b}{2} \right )\geq \frac{f(a)+f(b)}{2} \tag{2}$$

Inductively we can show that $(2)$ takes its equivelant form

$$f\left ( \frac{1}{n}\sum_{k=1}^{n}a_k \right )\geq \frac{1}{n}\sum_{k=1}^{n}a_k \tag{3}$$

Applying $(3)$ to the log. function we get that:

$$\begin{aligned}
\log \left ( \frac{1}{n}\sum_{k=1}^{n}a_k \right) &= \frac{1}{n} \sum_{k=1}^{n}\log a_k \\
&=  \frac{1}{n} \log \prod_{k=1}^{n}a_k \\
 &= \log \left ( \sqrt[n]{\prod_{k=1}^{n}a_k} \right ) 
\end{aligned}$$

Since $\log$ is strictly increasing on $(0, +\infty)$ the identity we want follows immediately from the last equation. Hence we proved the AM - GM  inequality, that is equation $(1)$.