# Proof by induction for inequalities

• Easty
In summary, the conversation discusses how to prove that a sequence, defined by X1=3 and Xn+1= (6Xn+1)/(2Xn+5) for all n\in N, is always greater than 0 for all values of n. The conversation presents a solution using induction and shows the necessary steps to prove this statement.

## Homework Statement

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

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

## 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

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

For the induction step, you have to show that xk + 1 > 1.

xk + 1 = (6xk + 1)/(2xk + 5)

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

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.