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The sequence of real numbers $a_1$, $a_2$, $a_3$...is such that $a_1$ $=$ $1$ and $a_{n+1} = (a_n + \frac{1}{a_n} )^{\lambda}$ ,where $\lambda$ is a constant greater than 1.

Prove by mathematical induction that for n ≥ 2,
$a_n$ ≥ $2^{g(n)}$ where $g(n)=\lambda^{n-1}$

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

This question is really confusing me. I cant even prove the base case

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Since you're asked to prove a property for $n\ge 2$, the base case is the $n=2$ case: $$a_2\ge 2^{g(2)}$$ This is very easy (in fact trivial) to prove once you unfold the defintions of $a_2$ and $g(2)$.

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  • $\begingroup$ The working simplifies to $2^\lambda \ge 2^\lambda$ $\endgroup$ – user140161 Sep 15 '14 at 20:14
  • $\begingroup$ @user140161: Exactly. And that is trivially true. $\endgroup$ – Henning Makholm Sep 15 '14 at 20:15
  • $\begingroup$ I've assumed it to be true n=k. Having trouble proving it for n=k+1 $\endgroup$ – user140161 Sep 15 '14 at 20:18
  • $\begingroup$ @user: Note that $a_n+\frac1{a_n}\ge a_n$, and ${-}^\lambda$ is strictly increasing because $\lambda>2$. The rest is just calculation. $\endgroup$ – Henning Makholm Sep 15 '14 at 20:21
  • $\begingroup$ Assuming the statement to be true for n=k, I got the inequality $a_k \ge 2^{\lambda^{k-1}}$. Then plugging in n as k+1 I got $a_{k+1} \ge 2^{\lambda^{k}}$ which I simplified to $(a_k + \frac{1}{a_k})^\lambda \ge 2^{\lambda^{k}}$. How do I prove this? $\endgroup$ – user140161 Sep 15 '14 at 22:22

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