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