TITLE: Prove $(x+r_1) \cdots (x+r_n) \geq (x+(r_1 \cdots r_n)^{1/n})^{n}$.
QUESTION [7 upvotes]: I can show that for $x > 0$ and $r_{i} > 0$ we have
$$
\left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\
\geq\
\left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.$$
However, I can't do this using straight up induction, strong or weak. Can someone do this? 
You can show the inequality holds for $n$ if it holds for $n+1$ using $r_{n+1} = (r_1 \cdots r_n)^{1/n}$, so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for $n=a$ and $n=b$ then it holds for $n=ab$. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal".
You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of $x^k$ and use the AM-GM inequality.

REPLY [2 votes]: Perhaps the simplest one is to use AM-GM only twice: $$n =  \sum_{k=1}^n\dfrac{x}{x+r_k}+\sum_{k=1}^n\dfrac{r_k}{x+r_k}\geq n\dfrac{x}{(\prod_{k=1}^n(x+r_k))^{\frac{1}{n}}}+n\dfrac{(r_1r_2...r_n)^{\frac{1}{n}}}{(\prod_{k=1}^n(x+r_k))^{\frac{1}{n}}},$$
which then will yield the original inequality immmediately.