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Prove by mathematical induction:

($\frac{a_{1} + ... + a_{n}}{n}$) $\geq \sqrt[\leftroot{0}\uproot{0}n]{a_{1}...a_{n}}$, where $n \in \mathbb{N}$.

So the base case: $n = 2$ means $\frac{a_{1} + a_{2}}{2} \geq a_{1}a_{2}$. Rearranging yields $(a_{1} - a_{2})^{2} \geq 0$.

The inductive step is where I am stuck at.

Any advice?

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Let $$G = \sqrt[n]{x_1 \cdots x_{n}}$$ and it then follows that $x_1 \leq G \leq x_n$. Also $$x_1 +x_n \geq \frac{x_1x_n}{G}+G.$$

Implies $$\frac 1G (G-x_1)(x_n - G) \geq 0.$$

By Induction Hypothesis:

$$x_2 + \cdots +x_{n-1} + \frac{x_1x_n}{G} \geq (n-1)G$$ $\implies$ $$x_2 + \cdots +x_{n-1} + \frac{x_1x_n}{G} +G \geq nG$$ and use $x_1 +x_n \geq \frac{x_1x_n}{G}+G$ to get the reqd result.

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  • $\begingroup$ You’re implicitly assuming $x_i$ are ordered (WLOG) and positive, probably worth stating. +1 $\endgroup$ – Macavity Oct 24 '17 at 4:44
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The idea of the induction using can be the following.

For $n=2^k$, where $k$ is a natural number we can use AM-GM number times, like for $n=4$:

For non-negatives $a$, $b$, $c$ and $d$ we obtain:

$$\frac{a+b+c+d}{4}\geq\frac{2\sqrt{ab}+2\sqrt{cd}}{4}=\frac{\sqrt{ab}+\sqrt{cd}}{2}\geq\sqrt{\sqrt{ab}\cdot\sqrt{cd}}=\sqrt[4]{abcd}.$$

Now, for $n=3$ we obtain: $$\frac{a+b+c+\sqrt[3]{abc}}{4}\geq\sqrt[4]{abc\sqrt[3]{abc}}=\sqrt[3]{abc},$$ which gives $$\frac{a+b+c}{3}\geq\sqrt[3]{abc}.$$

Or for non-negatives $a$, $b$, $c$, $d$ and $e$ we obtain: $$\frac{a+b+c+d+e+3\sqrt[5]{abcde}}{8}\geq\sqrt[8]{abcde\left(\sqrt[5]{abcde}\right)^3}=\sqrt[5]{abcde},$$ which gives $$\frac{a+b+c+d+e}{5}\geq\sqrt[5]{abcde}.$$

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