Solution to AM - GM Introductory Problem 1


For nonnegative real numbers $a_1,a_2,\cdots a_n$, demonstrate that if $a_1a_2\cdots a_n=1$ then $a_1+a_2+\cdots +a_n\ge n$.


Since $a_1a_2\cdots a_n=1$, the geometric mean ($\sqrt[n]{a_1a_2\cdots a_n}$) must also equal $1$.

The AM-GM Inequality states that the arithmetic mean of a set of non-negative numbers is greater than or equal to the geometric mean, so that means that $\frac{a_1+a_2+\cdots +a_n}{n}\geq 1$.

Rearranging, we get $a_1+a_2+\cdots +a_n\ge n$, as required. $\square$

Back to the Arithmetic Mean-Geometric Mean Inequality.