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I was able to prove by induction on $n\in\mathbb{N}$ that for any $m:=2^n$ positive real numbers we have $\text{Geometric Average}\leq\text{Arithmetic Average}$,

I.e. $\sqrt[m]{\Pi_{i=1}^m a_i} = $$\sqrt[2^n]{\Pi_{i=1}^{2^n} a_i} \leq\frac{\Sigma_{i=1}^{2^n} a_i}{2^n}$$ =\frac{\Sigma_{i=1}^m a_i}{m}$ where $a_1,...,a_m\in\mathbb{R}_+$ are any $m=2^n$ positive real numbers.

Now I am trying to prove by using this fact, That for any $n\in\mathbb{N}$ positive real numbers we have $\text{Geometric Average} \leq \text{Arithmetic Average}$,

I.e. $\sqrt[n]{\Pi_{i=1}^n a_i} \leq\frac{\Sigma_{i=1}^n a_i}{n}$ where $a_1,...,a_n\in\mathbb{R}_+$ are any $n$ positive real numbers, But then I got stuck and was not able to proceed.

In other words, The problem can be stated as:

Suppose that we know $\forall n\in\mathbb{N},\forall a_1,...,a_{2^n}\in\mathbb{R}_+, \sqrt[2^n]{\Pi_{i=1}^{2^n} a_i} \leq\frac{\Sigma_{i=1}^{2^n} a_i}{2^n}$

and prove by using this fact that $\forall n\in\mathbb{N},\forall a_1,...,a_{n}\in\mathbb{R}_+, \sqrt[n]{\Pi_{i=1}^{n} a_i} \leq\frac{\Sigma_{i=1}^{n} a_i}{n}$

Thanks for any hint/help...

($\mathbb{R}_+$ - The set of positive real numbers)


marked as duplicate by bof, Ennar, Robert Soupe, cansomeonehelpmeout, user284331 May 8 '18 at 18:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Have You already checked the Wiki? en.wikipedia.org/wiki/… $\endgroup$ – Logic_Problem_42 May 8 '18 at 4:34
  • $\begingroup$ @bof It seems to be a narrower case, but maybe in this particular instance everyone is best served by having this closed as a duplicate. $\endgroup$ – Robert Soupe May 8 '18 at 15:25

Assume that:

For any $k\in\mathbb{N}$ and for any $b_1,...,b_{2^k}\in\mathbb{R}_+$, we have that: \begin{align} \frac{b_1 + \cdots + b_{2^k}}{2^k} \geq \sqrt[2^k]{b_1 \cdots b_{2^k}} \tag{$\star$} \end{align}

We want to show that:

For any $n\in\mathbb{N}$ and for any $a_1,...,a_n\in\mathbb{R}_+$, we have that: $$ \frac{a_1 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 \cdots a_n} $$

To this end, choose any $n\in\mathbb{N}$ and choose any $a_1,...,a_n\in\mathbb{R}_+$. Since the sequence $\{2^i\}_{i=1}^\infty$ is unbounded above, we know that there exists some $k \in \mathbb N$ such that $n \leq 2^k$. Now define: $$ \mu = \frac{a_1 + \cdots + a_n}{n} $$ and for each $j \in \{1, \ldots, 2^k\}$, define: $$ b_j = \begin{cases} a_j &\text{if } j \leq n \\ \mu &\text{otherwise} \end{cases} $$ Then observe that: \begin{align*} \mu &= \frac{a_1 + \cdots + a_n}{n} \\ &= \frac{2^k(a_1 + \cdots + a_n)}{2^kn} \\ &= \frac{n(a_1 + \cdots + a_n) + (2^k - n)(a_1 + \cdots + a_n)}{2^k n} \\ &= \frac{(a_1 + \cdots + a_n) + (2^k - n)\mu}{2^k} \\ &= \frac{b_1 + \cdots + b_{2^k}}{2^k} \\ &\geq \sqrt[2^k]{b_1 \cdots b_{2^k}} &\text{by $(\star)$} \\ &= \sqrt[2^k]{(a_1 \cdots a_n)\mu^{2^k - n}} \end{align*}

Raising both sides to the power of $2^k$, we obtain: $$ \mu^{2^k} \geq (a_1 \cdots a_n)\mu^{2^k - n} $$ Multiplying both sides by $\mu^{n - 2^k}$, we obtain: $$ \mu^n \geq a_1 \cdots a_n $$ Finally, taking $n^{\text{th}}$ roots of both sides yields: $$ \frac{a_1 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 \cdots a_n} $$ as desired. $~~\blacksquare$


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