# Inequality harmonic and geometric mean

I'm trying to understand the proof on the bottom:

https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality

He uses the AM-GM inequality to prove the HG-GM-inequality, but I don't see how he manages to rewrite the sum like he did. Can anybody help me out on that?

• What sum?.... – Martín-Blas Pérez Pinilla Apr 15 at 10:14
• The sum after "The inequality ... is a direct consequence of AM-GM" – Julian Apr 15 at 10:19

Let $$y_i=\sqrt[n]{\frac{x_1\cdots x_n}{x_i^n}}=\frac{\sqrt[n]{x_1\cdots x_n}}{x_i}$$. Then by the AM-GM inequality,
\begin{align*} \frac{1}{n}\sum_iy_i&\ge\sqrt[n]{y_1\cdots y_n}\\ &=\sqrt[n]{\prod_i\frac{\sqrt[n]{x_1\cdots x_n}}{x_i}}\\ &=\sqrt[n]{\frac{x_1\cdots x_n}{x_1\cdots x_n}}\\ &=1 \end{align*}
As $$\sum_iy_i=\sqrt[n]{x_1\cdots x_n}\sum_i\frac{1}{x_i}$$, cross multiplying gives $$\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}$$
$$\root n\of{\frac{x_1\cdots x_n}{x_i^n}} = \frac{\root n\of{{x_1\cdots x_n}}}{\root n\of{x_i^n}} = \cdots$$