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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 $

Suggestions would be helpful.

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  • $\begingroup$ Can you find upper and lower bounds for ${x_{k+1}^2}$? $\endgroup$ – player3236 Oct 12 '20 at 1:18
  • $\begingroup$ @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 $\sqrt 2$. $\endgroup$ – mathstudent288 Oct 12 '20 at 1:25
  • $\begingroup$ I mean for some specific $k$, using $1 \le x_k^2 \le k$. $\endgroup$ – player3236 Oct 12 '20 at 1:26
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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 $1\le x_{n+1}\le\sqrt{n+1}$.

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  • $\begingroup$ Thank you so much! This is understandable but only intuitive for me after I have seen the solution. I wasn't expecting it to be so simple. $\endgroup$ – mathstudent288 Oct 12 '20 at 1:43

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