Lesson 7

Theorem (inequality of arithmetic and geometric means, AM–GM inequality). For any positive integer \(n\) and any \(x_{1}, \dotsc, x_{n} \in \mathbb{R}_{>0}\), \[\frac{x_1+x_2+\dotsb+x_n}{n}\ge \sqrt[n]{x_1x_2 \dotsb x_n}\ .\]

Proof. Let us scale the given numbers. That is, fix a positive constant \(\alpha\) and consider the following numbers: \[\begin{align*}y_{1}&=\alpha x_{1},\\ y_{2}&=\alpha x_{2},\\&\vdots\\ y_{n}&=\alpha x_{n}.\\\end{align*}\] When moving from \(x_{1}, \dotsc, x_{n}\) to \(y_{1}, \dotsc, y_{n}\), both the arithmetic mean and the geometric mean are multiplied by \(\alpha\): \[\begin{align*}\frac{y_1+\dotsb+y_n}{n}&= \alpha \cdot \frac{x_1+\dotsb+x_n}{n}\\ \sqrt[n]{y_1 \dotsb y_n}&= \alpha \cdot \sqrt[n]{x_1 \dotsb x_n}\end{align*}\] Hence, it suffices to prove the inequality for \(y_{1}, \dotsc, y_{n}\).

Now, fix \[\alpha=\frac{1}{\sqrt[n]{x_1 \dotsb x_n}}.\] Then, the product of \(y\)'s is \(1\): \[y_{1}\dotsb y_{n}=(\alpha x_{1})\dotsb (\alpha x_{n})=\alpha^{n} (x_{1}\dotsb x_{n})=\frac{1}{x_1\dotsb x_n}\cdot (x_{1}\dotsb x_{n})=1.\]

Thus, it suffices to prove the following statement:

for any positive integer \(n\) and any \(y_{1}, \dotsc, y_{n} \in \mathbb{R}_{>0}\) such that \(y_{1} \dotsb y_{n}=1\), \[y_{1}+\dotsb+y_{n} \ge n\ .\]