Bounded recursive sequence - Proof by induction 
Given the sequence $(x_n)$ defined by
\begin{cases}
x_1 &= 1\\ x_{n+1} &= \frac{1}{2}\left(x_n + \frac{2}{x_n}\right),
\end{cases}
prove that $1 \leq x_n \leq \frac{3}{2}, \forall n \in N$.

I verified the base case for $n=1$ and $n=2$.
Assumed that the boundaries hold for all $k \leq n$. Used the induction hypothesis to show that $1 \leq x_n \leq \frac{3}{2} \Rightarrow \frac{1}{2} \leq \frac{x_n}{2} \leq \frac{3}{4} $  and $1 \leq x_n \leq \frac{3}{2} \Rightarrow  \frac{2}{3} \leq \frac{1}{x_n} \leq 1$.
Adding term by term I got $1 \leq \frac{7}{6} \leq \frac{1}{2}(x_n + \frac{2}{x_n}) \leq \frac{7}{4} $.
Is my reasoning correct? How could I show that $\frac{3}{2}$ is also an upper boundary, since it is smaller than $\frac{7}{4}$?
PS: I found similar questions for which $x_n \leq \sqrt{2}$.

For induction step we have to prove 
$1\leqslant x\leqslant\frac32\Rightarrow 2\leqslant x+\frac2x\leqslant3$
that is $x^2-3x+2\leqslant0$ and $x^2-2x+2\geqslant0$ for $1\leqslant x\leqslant\frac32$
Can you finish? 
