Request for historical precedent of identity implying AM-HM inequality

Question

(That title's quite a mouthful, isn't it?)

A lot of my answers, though derived independently, are not new, and I often acknowledge this. However, I recently came up with a result that was new to me, so I am asking if it is really new.

My result came up while attempting to find an inductive proof of the inequality between the arithmetic and harmonic means. This is the question:

How can I prove that $(a_1+a_2+\dotsb+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n})\geq n^2$

This is the inequality between the arithmetic and harmonic means.

I eventually came up with this identity, which I do not recall having seen before:

Let $s_n =u_nv_n $ where $u_n=\sum_{k=1}^n a_k, v_n= \sum_{k=1}^n \dfrac1{a_k} $.

Then

$s_{n+1} =(\sqrt{s_n}+1)^2+\dfrac1{\sqrt{a_{n+1}}}(\sqrt{u_n}-a_{n+1}\sqrt{v_n})^2 $.

Since $s_1 = 1$, this immediately shows that $\sqrt{s_{n+1}} \ge \sqrt{s_n}+1$ so that $\sqrt{s_n} \ge n$, $s_n \ge n^2$ and the condition for $s_{n+1} = s_n+1$ is $a_{n+1} =\sqrt{\dfrac{u_n}{v_n}} =\sqrt{\dfrac{\sum_{k=1}^n a_k}{\sum_{k=1}^n \dfrac1{a_k}}} $.

So, my question is "Is this identity new?"

Of course the answer is probably "No", but a reference would be nice.

Answer 1

There is a small error in your recurrence equation, it should be

$s_{n+1} =(\sqrt{s_n}+1)^2+\dfrac1{\color{red}{a_{n+1}}}\bigl(\sqrt{u_n}-a_{n+1}\sqrt{v_n}\bigr)^2$

or, equivalently,

$s_{n+1} =(\sqrt{s_n}+1)^2+\bigl(\dfrac{\sqrt{u_n}}{\sqrt{a_{n+1}}}-\sqrt{a_{n+1}}\sqrt{v_n}\bigr)^2$.

This is a special case of the Brahmagupta–Fibonacci identity $$ (a^2 + b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2 $$ applied to $$ (a, b, c, d) = (\sqrt{u_n}, \sqrt{a_{n+1}}, \sqrt{v_n}, \dfrac 1{\sqrt{a_{n+1}}}) \, . $$