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The sequence of real numbers $a_1, a_2, a_3,...$ is such that $a_1=1$ and $$a_{n+1}=\left(a_n+\frac1{a_n}\right)^\lambda,$$ where $\lambda$ is a constant greater than $1$. Prove by mathematical induction that, for $n\ge2$, $$a_n\ge2^{g(n)},$$ where $g(n)=\lambda^{n-1}\tag{6}.$

Prove also that, for $n\ge 2$, $\frac{a_{n+1}}{a_n}>2^{(\lambda-1)g(n)}\tag{3}.$

I have trouble with the method of proving the inequality given. I have already proven that the base case where $n=2$ is true. But beyond that, I'm not sure how to go on.

  • $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Martin Sleziak Jan 14 '15 at 16:01

Since $a_n \geq 2^{g(n)} \to a_n+\dfrac{1}{a_n} \geq a_n \geq 2^{g(n)}\to a_{n+1} = \left(a_n+\dfrac{1}{a_n}\right)^{\lambda} \geq 2^{g(n)\cdot \lambda} = 2^{\lambda\cdot \lambda^{n-1}} = 2^{\lambda^n} = 2^{g(n+1)}$.

Also, $a_n+\dfrac{1}{a_n} > a_n \Rightarrow a_{n+1} > a_n^{\lambda} \Rightarrow \dfrac{a_{n+1}}{a_n} > a_n^{\lambda-1} \geq (2^{g(n)})^{\lambda-1}=2^{(\lambda-1)\cdot g(n)}$

  • $\begingroup$ Do you mind explaining why $a_n+\dfrac{1}{a_n} \geq 2^{g(n)}$ ? $\endgroup$ – space monkeys Jan 14 '15 at 15:23
  • $\begingroup$ See my new edit... $\endgroup$ – DeepSea Jan 14 '15 at 15:25

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