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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 $N$.

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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:

  1. As a function, $G(x_1,\ldots,x_N)$ is increasing in each individual argument $x_k$.
  2. If one split $x_1,x_2,\ldots,x_N$ into two groups, we have $$G(x_1,x_2,\ldots,x_N)^N = G(x_1,\ldots,x_M)^M G(x_{M+1},\ldots,x_N)^{N-M}$$
  3. In particular, if $N = 2M$ is even, this leads to $$G(x_1,x_2,\ldots,x_N) = G(G(x_1,\ldots,x_M),G(x_{M+1},\ldots,x_N))$$

For any $n \ge 2$, let $S_n$ be the statement

For any $(a_1,\ldots,a_n), (b_1,\ldots,b_n) \in \mathbb{R}_{+}^n$, $$ G(\{a_i + b_i\}) = \left(\prod_{k=1}^n (a_k+b_k)\right)^{1/n} \ge G(\{a_i\}) + G(\{b_i\}) = \left(\prod_{k=1}^n a_k \right)^{1/n} + \left(\prod_{k=1}^n b_k \right)^{1/n}$$

  • $S_2$ 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 $b_i = b$.

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  • $\begingroup$ Very nice, thank you. It may be a bit far fetched, but do you think this can be extended to two sets of positive reals, and then $G(\{b_i+a_i\}) \geq G(\{b_i\}) + G(\{a_i\})$ ? Obviously, the question here is a special case of that. $\endgroup$
    – Andreas
    Oct 28 '16 at 6:51
  • $\begingroup$ Thank you for the extended proof. Upon searching for "superadditivity of the geometric mean" one finds related material. However I realized that only later. $\endgroup$
    – Andreas
    Oct 29 '16 at 20:15
  • $\begingroup$ The generalization is also the subject of this : math.stackexchange.com/questions/29357/… $\endgroup$
    – Arnaud D.
    Apr 10 '19 at 22:20
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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.

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