AM-GM Inequality

The AM-GM inequality states that for $x_1, x_2, ..., x_n >0$ and $\lambda_1, \lambda_2, ..., \lambda_n \ge 0$ , such that $\sum_{i = 1}^n \lambda_i = 1$ , we have

\prod_{i} x_i^{\lambda_i} \le \sum_i \lambda_i x_i

If particular, when $\lambda_i = \frac{1}{n}$ , the inequality becomes

\sqrt[^n]{\prod_i x_i} \le \frac{\sum_i x_i}{n}

Replaying $x_i = \frac{1}{y_i}$ , we obtain

\frac{n}{\sum_i \frac{1}{y_i}} \le \sqrt[^n]{\prod_i y_i}

Another famous case is $n = 2$ and $\lambda_1 = \frac{1}{p}$ , $\lambda_2 = \frac{1}{q}$ and $x_1 = x^p$ , $x_2 = y^q$ , we get the Young's inequality

xy = (x^p)^\frac{1}{p} (y^q)^\frac{1}{q} \le \frac{x^p}{p} + \frac{y^q}{q}

The proof of AM-GM inequality is rather simple and follows immediately from the concavity of $\log$ function:

\sum_i \lambda_i \log x_i \le \log \sum_i \lambda_i x_i