Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is this relation true ?

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

Thank you

share|improve this question
2  
Use the concavity of $\log$. –  Philippe Malot Oct 18 '13 at 16:47

1 Answer 1

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 $\log$?)

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.