# And inequality on sums and products

Having trouble understanding why this passage is true: $$\left( \sqrt{\prod_{i=1}^n x_i} \, \right)^{-1} \le \frac 1 n \sum_{i=1}^n \frac 1 {x_i} \Longrightarrow \sqrt[n]{\prod_{i=1}^ n x_i} \ge \frac n {\sum_{i=1}^n \frac 1 {x_i}}$$ Can anyone help?

• In general the harmonic mean is smaller or equal than the geometric mean. Sep 25, 2017 at 19:07

Don't be confused by the messy expressions. What you start with is $$a^{-1}\le \frac{1}{c}b$$ Here $a$ is the expression with the $n$-th root and the product, $b=n$, and $c$ is the sum with the $\frac{1}{x_i}$.

We then multiply both sides of the inequality by $\frac{ac}{b}$, which is positive (since all three numbers are positive) to get $$\frac{c}{b}\le a$$

• Why make this more complicated that it is. It says that if $A\le B$ then $\dfrac 1 A \ge \dfrac 1 B,$ which is true if $A$ and $B$ are both positive. Sep 25, 2017 at 19:12

The arrow $\Longrightarrow$ means "If . . . then . . .". So this says $$\text{If } A \le B \text{ then } \frac 1 A \ge \frac 1 B. \tag 1$$ That is true if $A$ and $B$ are both positive.

The first inequality is says the geometric mean of $\dfrac 1 {x_1}, \ldots, \dfrac 1 {x_n}$ is less than or equal to the arithmetic mean of those same numbers. That is true if all the numbers are positive.

Then line $(1)$ above gets applied.

The result is the inequality that says the geometric mean, not of $\dfrac 1 {x_1}, \ldots, \dfrac 1 {x_n},$ but rather of $x_1,\ldots,x_n,$ is at least as big as the harmonic mean of those same numbers.
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