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**[SOLVED] I think of it as hard proof by induction**

## Homework Statement

given for the sequence [itex]a_1,a_2,a_3,a_4,...[/itex],[itex]a_1=1[/itex] and that

[tex]a_{n+1}=(a_n+\frac{1}{a_n})^{\lambda}[/tex]

where[itex]\lambda>1[/itex].Prove by mathematical induction that,for [itex]n \geq 2[/itex]

[tex]a_n \geq 2^{g(n)}[/tex]

where [itex]g(n)=\lambda^{n-1}[/itex]

Prove also,for [itex]n \geq 2[/itex]

[tex]\frac{a_{n+1}}{a_n}>2^{(\lambda -1)g(n)}[/tex]

## Homework Equations

## The Attempt at a Solution

well what I tried to do was to consider [itex]a_{n+1}-2^{g(n)}[/itex]

which after simplification gives:

[tex] \frac{(a_n^2+1)^{\lambda}}{a_n^{\lambda}}-2^{g(n)}[/itex]

which I then said should be greater than or equal to zero since the fraction on the left is +ve and 2^g(n) is positive since lambda is >1. Is this a good way to start?