TITLE: Shifted geometric mean
QUESTION [6 upvotes]: Let $b$ and $a_i$ be positive real numbers, $i= 1 \cdots N$.  Let the geometric mean $G(\{x_i\}) = \sqrt[N]{\prod_{i=1}^N{x_i}}$. Prove for a shift $b$, that $$G(\{b+a_i\}) \geq b + G(\{a_i\})$$
I.e. prove $\sqrt[N]{\prod_{i=1}^N{(b+a_i)}} \geq b + \sqrt[N]{\prod_{i=1}^N{a_i}}$.
I came across this problem while aiming to find alternative solutions to inequalities like this one, i.e. inequalities which contain not so easy shifts, like an additional summand $1 + \cdots$ in the denominator. Geometric averages occur in inequalities frequently, not least after homogenizing. So I wonder if there is a general inequality like the one asked here. If it's known in the literature, a reference would of course be fine (I couldn't find any).
What I attempted: I obviously took the inequality to the  power $N$ and checked inequalities for the individual terms arising. This works fine for small $N$ with AM-GM, but I couldn't figure out how this generalizes to arbitrary $N$.

REPLY [2 votes]: Note that the desired inequality is equivalent to the inequality $\prod\limits_{i=1}^{n}{(b+a_i)}\ge\left(b+\left(\prod\limits_{i=1}^{n}{a_i}\right)^{1/n}\right)^n$. Expanding the LHS gives
$$\prod\limits_{i=1}^{n}{(b+a_i)} = \sum\limits_{i=0}^{n}{\left(\sum\limits_{\text{cyc}}{\prod\limits_{k=1}^{i}{a_{j_k}}}\right)b^{n-i}} $$
where for $0\le i\le n$ the cyclic sum is taken over all combination of $i$ integers $1\le j_1<\dots<j_i\le n$. By the binomial identity, expanding the RHS gives
$$\left(b+\left(\prod\limits_{i=1}^{n}{a_i}\right)^{1/n}\right)^n = \sum\limits_{i=0}^{n}{{n\choose i}\left(\prod\limits_{j=1}^{n}{a_j}\right)^{i/n}b^{n-i}}.$$
Now, notice that there are $n\choose i$ combination of $i$ integers, and if we take the product of $\prod\limits_{k=1}^{i}{a_{j_k}}$ over all combinations of $i$ integers $1\le j_1<\dots< j_i\le n$, we would get a product of $i{n\choose i}$ numbers, and by symmetry each $a_i$ is represented $\frac{i{n\choose i}}{n}$ times. That is,
$$\prod\limits_{\text{cyc}}{\prod\limits_{k=1}^{i}{a_{j_k}}} = \left(\prod\limits_{j=1}^{n}{a_j}\right)^{i{n\choose i}/n}. $$
Hence, by the AM-GM inequality, for each $1\le i\le n$, we have
$$\left(\sum\limits_{\text{cyc}}{\prod\limits_{k=1}^{i}{a_{j_k}}}\right)/{n\choose i}\ge\left(\prod\limits_{\text{cyc}}{\prod\limits_{k=1}^{i}{a_{j_k}}}\right)^{1/{n\choose i}} = \left(\prod\limits_{j=1}^{n}{a_j}\right)^{i/n}. $$
(The result is true for $i=0$ as well if we interpret an empty product as $1$.) This implies that
$$\sum\limits_{\text{cyc}}{\prod\limits_{k=1}^{i}{a_{j_k}}}\ge{n\choose i}\left(\prod\limits_{j=1}^{n}{a_j}\right)^{i/n}$$
for all $i$, and hence
$$\prod\limits_{i=1}^{n}{(b+a_i)} = \sum\limits_{i=0}^{n}{\left(\sum\limits_{\text{cyc}}{\prod\limits_{k=1}^{i}{a_{j_k}}}\right)b^{n-i}}\ge \sum\limits_{i=0}^{n}{{n\choose i}\left(\prod\limits_{j=1}^{n}{a_j}\right)^{i/n}b^{n-i}} = \left(b+\left(\prod\limits_{i=1}^{n}{a_i}\right)^{1/n}\right)^n $$
i.e. $\prod\limits_{i=1}^{n}{(b+a_i)}\ge\left(b+\left(\prod\limits_{i=1}^{n}{a_i}\right)^{1/n}\right)^n$, as desired.