# Chrystal's proof of the arithmetic-mean-geometric-mean inequality

I was looking through Chrystal's "Algebra, Part II" (1900), available for free on the web, and I noticed that it had a proof of the arithmetic-mean-geometric-mean inequality by induction.

I have written it here in more modern notation.

This is not a question - I would like to make this part of the wiki. How do I do that?

I find it interesting that this proof uses the product terms rather than the sum terms as I did in Proof by induction of AM-GM inequality

Chrystals's proof, page 46, rewritten.

Suppose $$\frac1{n}\sum_{k=1}^n a_k \ge (\prod_{k=1}^n a_k)^{1/n}$$ or $$\sum_{k=1}^n a_k \ge n(\prod_{k=1}^n a_k)^{1/n}$$.

Then $$\sum_{k=1}^{n+1} a_k \ge n(\prod_{k=1}^n a_k)^{1/n}+a_{n+1}$$.

If we can show that $$n(\prod_{k=1}^n a_k)^{1/n}+a_{n+1} \ge (n+1)(\prod_{k=1}^{n+1} a_k)^{1/(n+1)}$$, we are done.

$$n(\prod_{k=1}^n a_k)^{1/n}+a_{n+1} \ge (n+1)(\prod_{k=1}^{n+1} a_k)^{1/(n+1)}$$ is the same as, dividing by $$a_{n+1}$$, $$n(\prod_{k=1}^n a_k/a^n_{n+1})^{1/n}+1 \ge (n+1)(\prod_{k=1}^{n+1} a_k/a^{n+1}_{n+1})^{1/(n+1)} = (n+1)(\prod_{k=1}^{n} a_k/a^n_{n+1})^{1/(n+1)}$$.

Letting $$z =(\prod_{k=1}^{n} a_k/a^n_{n+1})^{1/(n(n+1))}$$, this is $$nz^{n+1}+1 \ge (n+1)z^n$$ or $$(n+1)(z^{n+1}-z^n) \ge z^{n+1}-1$$ or $$(n+1)z^n(z-1) \ge z^{n+1}-1$$.

If $$z = 1$$ this is $$0 \ge 0$$ which is true.

If $$z > 1$$ this is $$(n+1)z^n \ge \dfrac{z^{n+1}-1}{z-1} =\sum_{k=0}^n z^k$$ which is true since $$z^n \ge z^k$$.

If $$0 < z < 1$$, since $$z-1 < 0$$, dividing by $$z-1$$ reverses the inequality, so this is $$(n+1)z^n \le \dfrac{z^{n+1}-1}{z-1} =\sum_{k=0}^n z^k$$ which is true since $$z^n \le z^k$$.