Proof of Inequality X_n > sqrt(n) by Induction

[acynicsmind]

Homework Statement

Show the sequence {X_n} defined by X_1 = 1, X_n+1 = X_n + 1/X_n for n > 1 obeys the inequality X_n > sqrt(n) for all n >= 2.

Homework Equations

The Attempt at a Solution

For n = 2, 2 > sqrt(2) and for n = 3, 2.5 > sqrt(2.5). So it holds so far. What to do here after initial step for the inductive step is confusing me. I've never seen an induction proof with an inequality before today and the ones I've looked at online are not helpful at all.

Any help is appreciated. Thank you.

 

[rock.freak667]

Well what I do in induction questions with inequalities, I just try to make up the formula for x_N+1 after assuming true for n=N.

So I'd invert both sides and then go from there.

 

[Mark44]

You haven't said exactly what you have done for the induction hypothesis and induction step, so here's what's left to do.
Assume that x_k > \sqrt{k} is true.
Show that x_{k + 1} > \sqrt{k + 1}. Be sure to use the given information that x_{k + 1} = x_k + 1/x_k.

 

[acynicsmind]

You haven't said exactly what you have done for the induction hypothesis and induction step, so here's what's left to do.
Assume that x_k > \sqrt{k} is true.
Show that x_{k + 1} > \sqrt{k + 1}. Be sure to use the given information that x_{k + 1} = x_k + 1/x_k.

See, it's the recursive formula that is kind of tripping me up because I am not sure where to use it. Do I simply sub it in for X_k when trying to show x_{k + 1} > \sqrt{k + 1} and then work from there?

 

[Mark44]

You know that x_{k + 1} = x_k + 1/x_k. Your induction hypothesis gives you something you can replace x_k with.