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Let $\ $ ${y_1\cdot...\cdot y_n} \in \mathbb{R}$ $\ $ be positive $\quad$

Prove: $\sqrt[n]{y_1\cdot...\cdot y_n}$ $\le$ $\frac{y_1+...+y_n}{n}$

I have tried to find this by searching keywords like gemometric and arithmetic average, but hadn't found.

Actually, I have no idea how to start proving this one.

Batominovski
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HelpMe
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  • I didn't understand why there is need to to go through all those steps to prove the formula I uploaded? – HelpMe Nov 11 '18 at 08:08
  • Actually, I have to prove it by using the following: ($0 < x_1,...,x_n \in \mathbb{F} \wedge x_1 \cdot ... \cdot x_n=1$) $\implies$ $x_1 + ... + x_n \ge n$ – HelpMe Nov 11 '18 at 08:09

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