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 $N$.
Best Answer
OP is curious about whether a generalization is possible in comment. Instead of the original inequality in question, I'm going to show a generalized version of that where $b$ can depend of $i$.
For any $x_1, \ldots, x_N \in \mathbb{R}_{+}$, let $G(x_1,x_2,\ldots,x_N) = \left(\prod\limits_{k=1}^N x_k\right)^{1/N}$ be their geometric mean.
We recall following properties of the geometric mean:
For any $n \ge 2$, let $S_n$ be the statement
$S_2$ is true.
Apply Cauchy Schwarz to $(\sqrt{a_1},\sqrt{b_1}), (\sqrt{a_2},\sqrt{b_2})$, we get $$ \sqrt{(a_1+b_1)(a_2+b_2)} = \sqrt{\left(\sqrt{a_1}^2+\sqrt{b_1}^2\right)\left(\sqrt{a_2}^2+\sqrt{b_2}^2\right)} \ge \sqrt{a_1a_2} + \sqrt{b_1b_2} $$ This is precisely $S_2$.
$S_2 \land S_n \implies S_{2n}$.
For any $\begin{align} (a_1,\ldots,a_{2n}) &= (a'_1,\ldots,a'_n,a''_1,\ldots,a''_n),\\ (b_1,\ldots,b_{2n}) &= (b'_1,\ldots,b'_n,b''_1,\ldots,b''_n) \end{align} \in \mathbb{R}_{+}^{2n} $, we have
$$\begin{array}{rll} G(\{ a_i + b_i \}) &= G(G(\{ a'_i + b'_i \}),G(\{a''_i + b''_i\})) & \color{blue}{\text{prop 3.}}\\ &\ge G(G(\{a'_i\}) + G(\{b'_i\}),G(\{a''_i\})+G(\{b''_i\})) & \color{blue}{S_n \text{ and prop 1.}}\\ &\ge G(G(\{a'_i\})G(\{a''_i\})) + G(G(\{b'_i\})G(\{b''_i\})) & \color{blue}{S_2}\\ &= G(\{a_i\}) + G(\{b_i\}) & \color{blue}{\text{prop 3.}}\\ \end{array} $$
By principle of induction, $S_n$ is true whenever $n = 2^k$ is a power of $2$.
For general $n > 2$ but not a power of two, let $k$ be the integer such that $2^{k-1} < n < 2^k$.
Let $\bar{a} = G(a_1,\ldots,a_n)$ and $\bar{b} = G(b_1,\ldots,b_n)$. Consider following two $2^k$-tuples: $$\begin{align} ( \tilde{a}_1,\ldots, \tilde{a}_{2^k}) &= ( a_1, a_2, \ldots, a_{n}, \bar{a}, \ldots, \bar{a} ),\\ ( \tilde{b}_1,\ldots, \tilde{b}_{2^k}) &= ( b_1, b_2, \ldots, a_{n}, \bar{b}, \ldots, \bar{b} ) \end{align} $$ It is easy to see $G(\{\tilde{a}_i\}) = \bar{a}$ and $G(\{\tilde{b}_i\}) = \bar{b}$.
Apply $S_{2^k}$ to the two $2^k$-tuples and raise both sides of result to $2^k$ power, we find
$$\begin{array}{rll} & G(\{ \tilde{a}_i + \tilde{b}_i \})^{2^k} \ge (G(\{\tilde{a}_i\}) + G(\{\tilde{b}_i\}))^{2^k}\\ \iff & G(\{a_i + b_i\})^n (\bar{a}+\bar{b})^{2^k - n} \ge (\bar{a}+\bar{b})^{2^k} & \color{blue}{\text{prop. 2 }}\\ \iff & G(\{a_i+b_i\}) \ge \bar{a} + \bar{b} = G(\{a_i\}) + G(\{b_i\}) \end{array} $$ This implies $S_n$ is true for $n$ other than a power of $2$ too.
As a result, $S_n$ is true for all $n \ge 2$.
The inequality in question is a special case of this where all $b_i = b$.