Proof by induction for inequalities

[Easty]

Homework Statement

A sequence (X_n) is defined by X₁=3 and X_n+1= (6X_n+1)/(2X_n+5) for all n$$\in$$ N.

Prove by induction or otherwise that X_n-1 > 0 for all n $$\in$$ N.

Homework Equations

The Attempt at a Solution

I'm not sure with what to do when dealing with inequalities in an induction proof. Initial i tried subing in the recursion formula when attempting the inductive step but i don't think it gets me anywhere. I'd really appreciate any guidance on where to start.

Thanks

 

[Mark44]

The proposition is clearly true for n = 1 and n = 2, so suppose it's true for n = k. I.e., that x_k > 1.

For the induction step, you have to show that x_{k + 1} > 1.

x_{k + 1} = (6x_k + 1)/(2x_k + 5)

Carry out the division to get something that you can show is greater than 1. Is that enough of a start?

 

[Russell Berty]

It is easier if you write it as x_n > 1 instead of x_n - 1 >0.

Suppose x_n > 1 [Show that x_(n+1) > 1]

Mult both sides by 4.
Add 2x to both sides.
Add 1 to both sides.
Divide both sides by the right side quantity (which is > 0, why?)

And you should see x_(n+1) > 1.

 

[Easty]

Thanks a lot for the help. Its such a simple solution, no wonder i didnt get it