Proof by induction, inequality

The sequence of real numbers $\ a_1,a_2,a_3.....$ is such that $\ a_1=1$ and

$$a_{n+1} = \left(a_n+\frac{1}{a_n}\right)^{\!\lambda}$$
where $\ \lambda >1$

Prove by mathematical induction that for $n\geq 2$

$$a_n\geq2^{g(n)}$$
where $g(n) = \lambda^{n-1}$

Attempt

I have solved the base case. I am having a problem in proving the statement for $n+1$. I tried two methods. Here they are:

Method 1

Assuming inductive hypothesis to be true

$$a_n\geq2^{g(n)}$$
$$a_n+\frac{1}{a_n}\geq2^{g(n)}$$
$$\left(a_n+\frac{1}{a_n}\right)^{\!\lambda}\geq(2^{g(n)})^{\!\lambda}$$
$$a_{n+1}\geq 2^{g(n+1)}$$

The statement holds true for $n+1$

Is this method correct?

Method 2

$$a_{n+1}-2^{g(n)} =a_{n+1}-2^{g(n)}$$
$$a_{n+1}-2^{g(n)} =\left(a_n+\frac{1}{a_n}\right)^{\!\lambda}-2^{g(n)}$$
$$a_{n+1}-2^{g(n)} =\frac{(a^2_n+1)^\lambda}{a_n^\lambda}-2^{g(n+1)}$$
$$a_{n+1}-2^{g(n)} =\frac{(a^2_n+1)^\lambda-(a_n^\lambda)(2^{g(n+1)})}{a_n^\lambda}$$
In this method, I don't know how to show $(a^2_n+1)^\lambda-(a_n^\lambda)(2^{g(n)})$ and $a_n^\lambda$ to be greater than or equal to $0$ which is necessary to prove the statement.

Can somebody provide me some hints which would prove beneficial in solving this problem?

Answer

Hint:

Step 1: Check that the conclusion holds for $n=1$ (which is immediate and necessary, but absent in your proof);

Step 2: Given that the conclusion holds for some $n\ge 1$ (i.e., $a_n\ge 2^{g(n)}$), show that it also holds for $n+1$, i.e., $a_{n+1}\ge 2^{g(n+1)}$. One way to show this is by first proving that the function

$$f(x)=\left(x+\frac 1x\right)^\lambda$$
is strictly increasing when $x\ge 1$, which is not difficult. With this result, we will see that

$$a_{n+1}=\left(a_n+\frac{1}{a_n}\right)^\lambda\ge \left(2^{g(n)}+2^{-g(n)}\right)^\lambda>2^{\lambda g(n)}=2^{g(n+1)}.$$
Completing the two steps yields the desired result.