Proof by induction if this is an increasing sequence
Mia Lopez 
Let, $$ x_{n+1}=\frac{x_{n}+\sqrt{3x_{n}}}{2}$$ be a sequence of real numbers, where $x_{1}=1 \space$. My goal is to prove that is a strict increase sequence, by induction. To be a strict increasing sequence, $x_{n+1} >x_{n}$ for all $n \in \mathbb{N}$. So, let $P(n)$ be the condition that claims $x_{n+1} >x_{n}$. For $P(1)$ one have:
$$\frac{1+\sqrt{3}}{2}>1 \Leftrightarrow 1+\sqrt{3}>2 \Leftrightarrow \sqrt{3}>1$$ Which is true. Now assume that $P(n)$ is true, lets prove $P(n+1)$. It is to say, $ \space x_{n+2}>x_{n+1}$.
$$\frac{x_{n+1}+\sqrt{3x_{n+1}}}{2}>\frac{x_{n}+\sqrt{3x_{n}}}{2}$$ $$x_{n+1}+\sqrt{3x_{n+1}}>x_{n}+\sqrt{3x_{n}}$$
If by the hypothesis $\space x_{n+1}>x_{n}$, $\space \sqrt{3x_{n+1}}>\sqrt{3x_{n}} \space$ would be true also. And, so $x_{n+1}+\sqrt{3x_{n+1}}>x_{n}+\sqrt{3x_{n}}$ it's true. One can conclude that $P(n+1)$ is true and the sequence is in fact strictly increasing.
Can you correct me, if I wrong. Thanks.
$\endgroup$ 52 Answers
$\begingroup$You might want to mention the fact that $f(a) = \sqrt{a}$ is a monotonically increasing function, that would justify your second step in the "induction step". Also, it should look something like this, $$ x_{n+1} > x_n \Rightarrow 3x_{n+1} > 3x_n \Rightarrow \text{ square root step } $$ and the "square root step" is justified by my prior comments. It looks good, I'm just being nit-picky. Overall, your proof is thorough.
$\endgroup$ $\begingroup$I wrote the reasoning below as a comment to another user, but it turned out to be unecessary as a comment, so I'm posting it as an answer:
Assume the Induction Hypothesis $x_{n+1}>x_n$ is true. Now consider the following sequence of implications: $$x_{n+1}>x_n \Longrightarrow x_{n+1}>x_n \wedge 3x_{n+1}>3x_n \Longrightarrow x_{n+1}>x_n \wedge \sqrt{3x_{n+1}}>\sqrt{3x_n}$$ $$\Longrightarrow x_{n+1} + \sqrt{3x_{n+1}}> x_n+ \sqrt{3x_n}\Longrightarrow P(n+1)$$
Since $x_{n+1}>x_n$ is true, it follows that $P(n+1)$ is true, as needed.
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