> [!NOTE] Theorem (Arithmetic/Geometric Mean Inequality) > Let $(x_{i})_{i}^{n}$ denote a finite [[Real sequences|sequence]] of positive real numbers. Then their geometric mean is less then or equal to their arithmetic mean i.e. $\left( \prod_{i=1}^{n} x_{i} \right)^{\frac{1}{n}} \leq \frac{1}{n} \sum_{i=1}^{n} x_{i}$with equality when $x_{1}=x_{2}=\dots=x_{n}$. **Proof**. Let $m=\frac{1}{n} \sum_{i=1}^{n} x_{i}.$ By [[Tangent to Real Natural Logarithm at x = 1]], $\begin{align} -\log m+ \frac{1}{n} \sum_{i=1}^{n} \log x_{i}&=\frac{1}{n} \sum_{i=1}^{n} \log \left( \frac{x_{i}}{m} \right) \\ & \leq \frac{1}{n} \left( \sum_{i=1}^{n} \left( \frac{x_{i}}{m } -1 \right) \right) \\ &= \frac{1}{n} \frac{1}{m} \left( \sum_{i=1}^{n} x_{i} \right) - 1 \\ &= \frac{m}{m}-1=0 \end{align}$hence $\frac{1}{n}\sum_{i=1}^{n} \log x_{i} \leq \log m$then $\log \left[ \left( \prod_{i=1}^{n} x_{i} \right)^{\frac{1}{n}} \right] \leq \log m$so $\left( \prod_{i=1}^{n} x_{i} \right)^{\frac{1}{n}} \leq m$since $\log$ is strictly increasing.