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 $z^n \le z^k$.



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