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I am working on the following proof and have gotten about half way;

If [tex]a_1, a_2, \ldots , a_n[/tex] are positive real numbers then

[tex] \sqrt[n]{a_1 \cdots a_n} \leq \frac{a_1+a_2+\ldots +a_n}{n}[/tex]

By induction, I started by showing it for n=2, which goes as follows.

We must show that [tex]\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}[/tex]. So consider some s, a positive real number. If

[tex]\sqrt{a_1 a_2} \leq s \Rightarrow \sqrt{a_1 a_2} \leq \frac{\sqrt{a_1a_2}+s}{2} \leq s[/tex]

And since [tex]\sqrt{a_1a_2}<a_1+a_2-\sqrt{a_1a_2}[/tex], just let that equal s. Then it follows that,

[tex]\sqrt{a_1a_2} \leq \frac{\sqrt{a_1a_2}+a_1+a_2-\sqrt{a_1a_2}}{2}[/tex]

Where after simplification we have shown that [tex]\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}[/tex] or in other words it is true for n=2.

Now for the next part of the proof, I think we must show that if it is true for [tex]n=2^m[/tex] then it is also true for [tex]n=2^{m+1}[/tex]. And then we must also show it if [tex]2^m < n < 2^{m+1}[/tex], but this is where I am stumped. Sorry for the choppiness in my explanation. Thanks for any assistance, I appreciate it.

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# An interesting inequality-question on the proof

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