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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?

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  • $\begingroup$ What sum?.... $\endgroup$ – Martín-Blas Pérez Pinilla Apr 15 at 10:14
  • $\begingroup$ The sum after "The inequality ... is a direct consequence of AM-GM" $\endgroup$ – Julian Apr 15 at 10:19
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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 $$\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}$$

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$$\root n\of{\frac{x_1\cdots x_n}{x_i^n}} = \frac{\root n\of{{x_1\cdots x_n}}}{\root n\of{x_i^n}} = \cdots$$

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