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Is this relation true ?

$\Pi_{i=1}^n v_n \le \left(\frac{\sum_{i=1}^{n} v_n}{n}\right)^n$

Thank you

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Use the concavity of $\log$. – Philippe Malot Oct 18 '13 at 16:47

This is true in some circumstances; if you rearrange things slightly, it is equivalent to assert $$ \left(\Pi_{i=1}^n v_i\right)^{1/n} \le \frac{\sum_{i=1}^{n} v_i}{n}. $$

Another way of saying this is 'the geometric mean of $(v_n)$ is less than the arithmetic mean of $(v_n)$'.

Now, as I suggested earlier, this isn't always true; if $n=2$, $v_1 = v_2 = -1$ for instance, it fails. I'll leave finding the correct hypothesis up to you, and a proof can then be given using, as girianshiido suggests, using $\log$. (The involvement of $\log$ also serves as a hint regarding the correct hypothesis - what is the domain of $\log$?)

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This is true for strictly positive $v_i$ ? – Dingo13 Oct 18 '13 at 17:03
Yeah, the $\log$ argument will show that this is true for strictly positive $v_i$. Then, if you assume that the $v_i$ are only non-negative, if any of the $v_i = 0$, you can demonstrate this directly. – BaronVT Oct 18 '13 at 17:05
In fact I have responded intuitively. By looking your slight rearrangement, I notice the concavity property of log but I don't notice a contradiction when I take a zero value for one $v_i$, the left term approaches $-\infty$ but not necessarily the right. So where I'm wrong ? – Dingo13 Oct 18 '13 at 17:26
The inequality still holds if one (or more) $v_i = 0$, just show it directly: the product of some non-negative numbers and $0$ on the left hand side vs. the sum of some non-negative numbers on the right. – BaronVT Oct 18 '13 at 17:50

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