# Use the arithmetic-geometric inequality for this list to deduce the arithmetic-geometric inequality for $n$.

Suppose that $n$ is not a power of two. Let $2^k$ be a power of $2$ that exceeds $n$ and consider the list $$a_1,\dots,a_n,\underbrace{A,A,\dots,A}_\text{2^k-n times}$$ of length $2^k$. Use the arithmetic-geometric inequality for this list to deduce the arithmetic-geometric inequality for $n$.

Proof: The arithmetic mean of the list above is \begin{equation*} \begin{aligned} & \frac{a_1+\dots+a_n+\overbrace{A+\dots+A}^\text{$2^k-n$ times}}{n+2^k-n} \\ = & \frac{a_1+\dots+a_n+\overbrace{\left(\frac{a_1+\dots+a_n}{n}\right)+\dots+\left(\frac{a_1+\dots+a_n}{n}\right)}^\text{$2^k-n$ times}}{2^k} \\ = & \frac{\dfrac{n(a_1+\dots+a_n)}{n}+\overbrace{\left(\frac{a_1+\dots+a_n}{n}\right)+\dots+\left(\frac{a_1+\dots+a_n}{n}\right)}^\text{$2^k-n$ times}}{2^k} \\ = & \frac{\dfrac{(n+2^k-n)(a_1+\dots+a_n)}{n}}{2^k} = \frac{\dfrac{(2^k)(a_1+\dots+a_n)}{n}}{2^k} = \frac{2^k(a_1+\dots+a_n)}{n}\frac{1}{2^k} = \frac{(a_1+\dots+a_n)}{n} \\ \geq & \sqrt{a_1\cdot \dots \cdot a_n} \end{aligned} \end{equation*}

I am unsure if I did what the question is asking of me.

• How do you justify the last passage? (Isn't it what you need to prove?) – Oskar Limka Feb 1 '16 at 21:22
• That is what I am confused about. I know the last part is true from a previous exercise. So is it ok to use that? – Username Unknown Feb 1 '16 at 21:34
• @UsernameUnknown It would help if you clarified what exactly was proved before, which you are allowed to use now. Was it maybe the AM-GM inequality, but just for the case where $n$ is a power of $2$? Also, recheck the last line you posted, that's not the GM of the $n$ numbers. – dxiv Feb 2 '16 at 0:06