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 $\log$?)
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 $\log$?)