# Monthly Archives: June 2016

## A new AM-GM inequality

I recently found the following inequalities useful:

$\displaystyle \left(x+\frac{a_1+\cdots+a_n}{n}\right)^n\ge (x+a_1)\cdots(x+a_n)\ge \left(x+\sqrt[n]{a_1\cdots a_n}\right)^n$,

for any $x,a_1,\dots,a_n\ge 0$.

To prove the left half simply apply AM-GM to the product

$\displaystyle (x+a_1)\cdots (x+a_n)\le \left(\frac{(x+a_1)+\cdots+(x+a_n)}{n}\right)^n$.

The right half follows directly from Hölder’s inequality.

More generally, one has

$\displaystyle\left(\frac{a_1+\cdots+a_n}{n}+\frac{b_1+\cdots+b_n}{n}\right)^n\ge (a_1+b_1)\cdots(a_n+b_n)\\\ge \left(\sqrt[n]{a_1\cdots a_n}+\sqrt[n]{b_1\cdots b_n}\right)^n$

for $a_1,\dots,a_n,b_1,\dots,b_n\ge 0$. The proof goes similarly.

One obtains similar inequalities for any number of vectors $a,b,c,\dots$.

As an example, for $a_i=i$ we get the inequalities

$\displaystyle \left(x+\frac{n+1}{2}\right)^n\ge (x+1)\cdots (x+n)>\left(x+\frac ne\right)^n$

using the fact that $\sqrt[n]{n!}/n\to 1/e$ from above.