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?

  • $\begingroup$ For part $(1)$, how about induction? Proposition $P(n)::=x_n\lt y_n$ $\endgroup$ – AnotherJohnDoe Oct 8 '17 at 17:45
  • $\begingroup$ If you have $x_n\lt y_n$, then can you compare $\sqrt{x_ny_n}$ to $y_n$? $\endgroup$ – AnotherJohnDoe Oct 8 '17 at 19:40
  • $\begingroup$ Using $P(n)$, show that $x_{n+1}\gt x_n$ $\endgroup$ – AnotherJohnDoe Oct 8 '17 at 19:46
  • $\begingroup$ To your first comment - yes. To your second, the idea is to show that $x_n\lt x_{n+1}\lt y_{n+1}\lt y_n$ $\endgroup$ – AnotherJohnDoe Oct 9 '17 at 5:27
  • $\begingroup$ Since we have $x_n\lt y_n$, we have $\sqrt{x_ny_n}\lt y_n$ $\endgroup$ – AnotherJohnDoe Oct 10 '17 at 5:02

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, $$ y_{n+1}-x_{n+1}=\sqrt{x_n y_n}-\frac{2}{{1\over x_n}+{1\over \sqrt{x_ny_n}}}\\ =\frac{\sqrt{y_n}-\sqrt{x_n}}{{1\over \sqrt{x_n}}+{1\over\sqrt{y_n}}}=\frac{y_n-x_n}{\left({1\over \sqrt{x_n}}+{1\over\sqrt{y_n}}\right)(\sqrt{y_n}+\sqrt{x_n})}<\frac{y_n-x_n}{2}. $$

  • $\begingroup$ @Sam you need to prove $x_n$ is monotone increasing, then you can say $x_n$ has a limit. Same for $y_n$, it decreases hence has a limit. Only the condition $y_n-x_n\to 0 $ can not convince these two sequences converge. $\endgroup$ – yahoo Oct 8 '17 at 23:55
  • $\begingroup$ @Sam Look at my induction, $x_{n-1}\leq x_n\leq y_n\leq y_{n-1}$, this just give the increasing and decreasing conditions $x_{n-1}\leq x_n$ and $y_n\leq y_{n-1}$. More explicitly, the first part i have proved is $x_{1}\leq x_{2}\leq\dots\leq x_{n-1}\leq x_{n}\leq\dots \leq y_{n}\leq y_{n-1}\leq\dots\leq y_{1}.$ Hence they are bounded and monotone. $\endgroup$ – yahoo Oct 9 '17 at 5:27
  • $\begingroup$ @Sam For $b\leq 2a$, I didn't use it, don;t know why you have this condition for your problem. $\endgroup$ – yahoo Oct 9 '17 at 5:34
  • $\begingroup$ @Sam $x_1-x_0=\frac{2}{{1\over a}+{1\over\sqrt{ab}}}-a>0$ iff $b>a$. It is just simple arithmatic, you can check it yourself for $y_n$. Any other questions about this answer? $\endgroup$ – yahoo Oct 9 '17 at 14:15
  • $\begingroup$ Okay, I actually understand how you got part (b) now as I have taken the time to attempt it myself. Now, from here I would need to use induction correct? $\endgroup$ – Sam Oct 11 '17 at 3:21

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