I recently found the following inequalities useful:
![\displaystyle \left(x+\frac{a_1+\cdots+a_n}{n}\right)^n\ge (x+a_1)\cdots(x+a_n)\ge \left(x+\sqrt[n]{a_1\cdots a_n}\right)^n](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%28x%2B%5Cfrac%7Ba_1%2B%5Ccdots%2Ba_n%7D%7Bn%7D%5Cright%29%5En%5Cge+%28x%2Ba_1%29%5Ccdots%28x%2Ba_n%29%5Cge+%5Cleft%28x%2B%5Csqrt%5Bn%5D%7Ba_1%5Ccdots+a_n%7D%5Cright%29%5En&bg=ffffff&fg=333333&s=0 "\displaystyle \left(x+\frac{a_1+\cdots+a_n}{n}\right)^n\ge (x+a_1)\cdots(x+a_n)\ge \left(x+\sqrt[n]{a_1\cdots a_n}\right)^n")
for any 
To prove the left half simply apply AM-GM to the product
The right half follows directly from Hölder’s inequality.
More generally, one has
![\displaystyle\left(\frac{a_1+\cdots+a_n}{n}+\frac{b_1+\cdots+b_n}{n}\right)^n\ge (a_1+b_1)\cdots(a_n+b_n)\\\ge \left(\sqrt[n]{a_1\cdots a_n}+\sqrt[n]{b_1\cdots b_n}\right)^n](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cleft%28%5Cfrac%7Ba_1%2B%5Ccdots%2Ba_n%7D%7Bn%7D%2B%5Cfrac%7Bb_1%2B%5Ccdots%2Bb_n%7D%7Bn%7D%5Cright%29%5En%5Cge+%28a_1%2Bb_1%29%5Ccdots%28a_n%2Bb_n%29%5C%5C%5Cge+%5Cleft%28%5Csqrt%5Bn%5D%7Ba_1%5Ccdots+a_n%7D%2B%5Csqrt%5Bn%5D%7Bb_1%5Ccdots+b_n%7D%5Cright%29%5En&bg=ffffff&fg=333333&s=0 "\displaystyle\left(\frac{a_1+\cdots+a_n}{n}+\frac{b_1+\cdots+b_n}{n}\right)^n\ge (a_1+b_1)\cdots(a_n+b_n)\\\ge \left(\sqrt[n]{a_1\cdots a_n}+\sqrt[n]{b_1\cdots b_n}\right)^n")
for 
One obtains similar inequalities for any number of vectors 
As an example, for 
using the fact that ![\sqrt[n]{n!}/n\to 1/e](https://s0.wp.com/latex.php?latex=%5Csqrt%5Bn%5D%7Bn%21%7D%2Fn%5Cto+1%2Fe&bg=ffffff&fg=333333&s=0 "\sqrt[n]{n!}/n\to 1/e")
