# Prove two statements by induction

Let $$x_1 = 1$$ and $$x_{n+1}$$ = $$\sqrt{{x_n}^2 + \frac{1}{(x_n)^2}}$$

Prove by mathematical induction that for all $$n ≥ 1, 1 ≤ x_n ≤ \sqrt n$$

I tested $$P(1)$$ and found $$1 \leq 1 \leq 1$$, which holds up.

Now I need to show $$P(n+1)$$.

By the inductive assumption, $$1 \leq x_k \leq \sqrt k$$ where $$k=n$$ and $$k \geq 1$$.

Then I did this, but I'm not sure what to do from here.

$$1 \leq x^2_k \leq k$$

• Can you find upper and lower bounds for ${x_{k+1}^2}$? – player3236 Oct 12 '20 at 1:18
• @player3236 The lower bound would be 2 where k = 1, and the upper bound would be infinity. That means the lowest possible value for $x_k+1$ would be $\sqrt 2$. – mathstudent288 Oct 12 '20 at 1:25
• I mean for some specific $k$, using $1 \le x_k^2 \le k$. – player3236 Oct 12 '20 at 1:26
The induction hypothesis is $$1\le x_n\le\sqrt n$$. Then $$1\le x_n^2\le n$$, so $$0<\frac{1}{x_n^2}\le 1$$. Therefore $$1\le x_n^2+\frac{1}{x_n^2}\le n+1$$, and finally $$1\le\sqrt{x_n^2+\frac{1}{x_n^2}}\le \sqrt{n+1}$$. So $$1\le x_{n+1}\le\sqrt{n+1}$$.