Solution to AM - GM Introductory Problem 1

Problem

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

Solution

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.

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Solution to AM - GM Introductory Problem 1

Problem

Solution