# Shifted geometric mean

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 $N$.

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 $$b_i = b$$.

• 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. Oct 28 '16 at 6:51
• Thank you for the extended proof. Upon searching for "superadditivity of the geometric mean" one finds related material. However I realized that only later. Oct 29 '16 at 20:15
• The generalization is also the subject of this : math.stackexchange.com/questions/29357/… Apr 10 '19 at 22:20

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.