Show that $G_{2^n}\le A_{2^n}$ by using induction on n.

I've proven the base case in the previous exercise:

Let $G_2=\sqrt{a_1a_2}$ and $A_2=\frac{1}{2}(a_1+a_2)$ and $a_1,a_2 \in \mathbb{R}$ $$\sqrt{a_1a_2}\le \frac{1}{2}(a_1+a_2)$$ $$2\sqrt{a_1a_2}\le (a_1+a_2)$$ $$4a_1a_2\le a_1^2+2a_1a_2+a_2^2$$ $$0\le a_1^2-2a_1a_2+a_2^2$$ $$0\le (a_1-a_2)^2$$

which is true for all real numbers.

Where I'm having the issue is adding the $(k+1)$ to the inequality.

For k: $G_{2^k} = (a_1a_2\cdots a_{2^{k}})^{\frac{1}{2^{k}}}$

For $2^{k+1}$:

$G_{2^{k+1}} = ((a_1a_2\cdots a_{2^{k}})^{\frac{1}{2^k}})^{\frac{k}{k+1}}\cdot a_{2^{k+1}}^{\frac{1}{2^{k+1}}}$

Am I on the right track here?


Hint: Apply the two term AM-GM to $$ (a_1a_2\dots a_n)^{\frac1n}\le\tfrac1n(a_1+a_2+\dots+a_n) $$ and $$ (a_{n+1}a_{n+2}\dots a_{2n})^{\frac1n}\le\tfrac1n(a_{n+1}+a_{n+2}+\dots+a_{2n}) $$

Note: you are only being asked to show this for groups of numbers whose size is a power of $2$.

  • $\begingroup$ And just like that it makes perfect sense. Thanks, robjohn. $\endgroup$ – russCam Feb 10 '14 at 22:34

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