Prove the AG inequality for integers $n$ that are powers of two, i.e. $n=2^k$.

Suppose $a_1,a_2,\dots,a_n>0$. The arithmetic mean of these numbers is $$\frac{a_1+\dots+a_n}{n}.$$ The geometric mean is $$ \sqrt{a_1\cdot \dots \cdot a_n}.$$ The arithmetic-geometric (AG) inequality states $$A\geq G .$$

Proof by induction: Let $k=0$. Then $$a_1\geq \sqrt{a_1} \iff \sqrt{a_1}\geq 0$$ which is true since $a_1>0$.

Induction step: Assume that $k=m$ is true. i.e. $$\frac{a_1+\dots+a_{2^k}}{2^k} \geq \sqrt{a_1\cdot \dots \cdot a_{2^k}}.$$ We need to show that this works for $k=m+1$. Well,

$$ \frac{a_1+\dots+a_{2^k}+a_{2^{k+1}}}{2^{k+1}} =\frac{1}{2}(\frac{a_1+\dots+a_{2^k}}{2^k})+\frac{a_{2^{k+1}}}{2^{k+1}} \geq \frac{1}{2}\sqrt{a_1\cdot \dots \cdot a_{2^k}} + \frac{a_{2^{k+1}}}{2^{k+1}} $$ How would I contunie from here?

  • $\begingroup$ Is the question to prove the AM-GM inequality for a total of $2^n$ integers, or an arbitrary number of intergers of the form $a_1 = 2^{n_1}, a_2 = 2^{n_2}, \dots$? $\endgroup$
    – anomaly
    Feb 1, 2016 at 20:45
  • $\begingroup$ In your last display, the passage is wrong, you need to add $a_{2^k+1}+\dotso+" in the first numerator, fix the second term in the second member, and possibly use the AG again to close in the third member? $\endgroup$ Feb 1, 2016 at 20:47

1 Answer 1


You missed some $a_n's$ in your induction step. And the geometric mean is supposed to be $\sqrt[n]{a_1a_2\cdots a_n}$.

Induction step: $$ {\sum_{n=1}^{2^k}a_n+\sum_{n=2^k+1}^{2^{k+1}}a_n\over 2^{k+1}}\ge{1\over2}\left(\sqrt[2^k]{\prod_{n=1}^{2^k}a_n}+\sqrt[2^k]{\prod_{n=2^k+1}^{2^{k+1}}a_n}\right)\ge\sqrt[2^{k+1}]{\prod_{n=1}^{2^{k+1}}a_n} $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .