My question reads:
For any positive $a,b$, the harmonic mean, $H(a,b)=\frac{2}{\frac{1} {a}+\frac{1} {b}}$ is less than or equal to the geometric mean $G(a,b)=\sqrt{ab}$.
This had to be proven, but I got this. The proof was quite straightforward.
(1)Given that $a,b$ are positive with $a<b$, define two sequences recursively by $x_0=a$, $x_{n+1}=H(x_n,y_{n+1})$ and $y_0=b$, $y_{n+1}=G(x_n,y_n)$. Prove that both sequences converge.
Thoughts: I am not too sure if I understand what exactly $x_{n+1}$ and $y_{n+1}$ are but I think it would be that $x_{n+1}=\frac{2}{\frac{1} {x_n}+\frac{1} {\sqrt{x_n\ y_n}}}$ and then $y_{n+1}=\sqrt{x_ny_n}$.
Then, I was thinking of using the Monotone Convergence Theorem, but I am having a bit of trouble determining the bound for each sequence so I can use the theorem. I am not sure how to begin the induction to show that the sequences are monotone increasing/decreasing to use the Theorem.
How can I use the fact that the Harmonic mean is less than or equal to the Geometric mean for part (1)?
(2) Prove that both sequences converge to the same number by considering $y_{n+1}-x_{n+1}$.
I am having trouble with this proof because I am not sure on what to do for part (1). Would induction be best in this case? I do not know how the assuming $b\leq\ 2a$ helps in this case.
Any suggestions?
To use induction, assume $x_{n-1}\leq x_n\leq y_n\leq y_{n-1}$. Then you can easily prove $x_n\leq x_{n+1}\leq y_{n+1}\leq y_{n}$. For example, to get $x_{n+1}\leq y_{n+1}$, $$ x_{n+1}=\frac{2}{{1\over x_n}+{1\over y_{n+1}}}=\frac{2}{{1\over x_n}+{1\over \sqrt{x_ny_n}}}\leq \frac{2}{{1\over \sqrt{x_ny_n}}+{1\over \sqrt{x_ny_n}}}=y_{n+1}. $$
For the second part, $$ y_{n+1}-x_{n+1}=\sqrt{x_n y_n}-\frac{2}{{1\over x_n}+{1\over \sqrt{x_ny_n}}}\\ =\frac{\sqrt{y_n}-\sqrt{x_n}}{{1\over \sqrt{x_n}}+{1\over\sqrt{y_n}}}=\frac{y_n-x_n}{\left({1\over \sqrt{x_n}}+{1\over\sqrt{y_n}}\right)(\sqrt{y_n}+\sqrt{x_n})}<\frac{y_n-x_n}{2}. $$