# I need help with induction

Given that

$$x_1=1$$ and $$\forall x \in \mathbb N :$$ $$x_{n+1} =$$ $$\sqrt{3+2x_n}$$

Use the principle of mathematical induction to prove, for all $$n \in \mathbb N$$, that $$x_n < x_{n+1}$$

So far I have done the base case and a little of the induction hypothesis, but I do not know how to proceed.

Base case: $$n=1$$

Then $$x_1 = 1$$ and $$x_2 = \sqrt{3+2(1)} = \sqrt{5}$$. Clearly, $$1<\sqrt{5}$$ , so this base step works.

Now, for my induction hypothesis I have $$\forall n \in \mathbb N, x_n < x_{n+1}$$ and I want to show that $$x_{n+1} < x_{n+2}$$

I start with: $$x_n < x_{n+1}$$ $$\to$$ $$x_n < \sqrt{3+2x_n}$$

Then $$x_n < \sqrt{3+2x_n}$$ $$\to$$ $$(x_n)^2 < 3+2x_n$$

So, $$\cfrac{(x_n)^2}{2} -3 < x_n$$

Here is where I came to a complete stop. Does anyone have suggestions on what I did wrong, or how to go from here? Also, sorry for any format issues. I am getting used to mathjax.

• Do you mean $x_{n+1}$ or $x_n+1$? Apr 10 '20 at 4:59
• "$q\lt q+1$" should not need induction. Surely you mean $x_n\lt x_{n+1}$ which is written x_n\lt x_{n+1}...? Apr 10 '20 at 5:00
• Yes I mean x_n\lt x_{n+1} Apr 10 '20 at 5:01

$$Hint:$$ Find some constant $$a\in \mathbb R$$ such that $$x_n \lt a$$ $$\forall n \in \mathbb N.$$

And using above fact complete your induction step by considering $$x_{n+1}^2-x_n^2$$

$$a=3$$ $$x_n<3 \Rightarrow (3+2x_n)<9 \Rightarrow x_{n+1}<3$$ $$x_{n+1}^2-x_n^2=3+2x_n-x_n^2=(3-x_n)(1+x_n)>0$$

You know that $$x_n \lt x_{n+1}$$

$$\implies x_n \lt \sqrt{3+2x_n}$$

$$\implies 2x_n \lt 2 \sqrt{3+2x_n}$$

$$\implies 3+ 2x_n \lt 3+ 2 \sqrt{3+2x_n}$$

$$\implies \sqrt{3+ 2x_n} \lt \sqrt{3+ 2 \sqrt{3+2x_n}}$$

$$\implies x_{n+1} \lt \sqrt{3+ 2 x_{n+1}}$$

$$\implies x_{n+1} \lt x_{n+2}$$

QED.