Show that $ x_{n+1}=\dfrac{4+3x_n}{3+2x_n} $is an increasing sequence? 
Let $(x_n)_{n \in \mathbb{N}}$ be the sequence defined by $x_1=1$ and $ x_{n+1}=\dfrac{4+3x_n}{3+2x_n} \text{for} \hspace{0.2 cm }n=1,2,3...$

Show that $(x_n)_{n \in \mathbb{N}}$ is an increasing sequence
First I used to prove $x_{n}\leq x_{n+1}$ by using mathematical induction but it is not success then I used a function
$f(x)=\dfrac{4+3x}{3+2x} $ which first derivative is always positive thus I conclude that $f(x)$ is an increasing function thus $x_{n}\leq x_{n+1}$
Can anyone help me I don't know what I am doing is correct or not?
If anyone help me to prove this using mathematical induction it is better!!
 A: You have differentiated the function and shown that it is positive, you can also check that $x_1<x_2$ manually. Then you have from the mean value theorem
$$x_{n+1}-x_n=\frac{4+3x_n}{3+2x_n}-\frac{4+3x_{n-1}}{3+2x_{n-1}}=f'(c)\cdot(x_n-x_{n-1})>0, c \in (x_{n-1},x_n)$$
where we used the induction hypothesis $x_n>x_{n-1}$ and the fact that the derivative is bigger than zero.
Only induction:
First use induction to show that $x_n$ is positive, this is very easy.
Then we get:
Check first manually that $x_1<x_2$. Assume now that $x_{n-1}<x_n$ we get
$$x_{n+1}-x_n=\frac{4+3x_n}{3+2x_n}-\frac{4+3x_{n-1}}{3+2x_{n-1}}\\=\frac{(4+3x_n)(3+2x_{n-1})-(4+3x_{n-1})(3+2x_n)}{(3+2x_n)(3+2x_{n-1})}\\=\frac{12+8x_{n-1}+9x_n+6x_nx_{n-1}-12-8x_n-9x_{n-1}-6x_nx_{n-1}}{(3+2x_n)(3+2x_{n-1})}\\=\frac{x_n-x_{n-1}}{(3+2x_n)(3+2x_{n-1})}>0,$$
where we have used the induction hypothesis, $x_n>x_{n-1}$ and the fact that $x_n,x_{n-1}$ is positive.
A: Without induction:
As all terms in this sequence are positive, we have
$$x_{n+1}=\frac{3x_n+4}{2x_n+3}>x_n\iff (2x_n+3)x_n<3x_n+4\iff 2x_n^2<4\iff x_n<2$$
Now rewrite the  recurrence relation:
$$x_{n+1}=\frac{3x_n+4}{2x_n+3}=\frac32+\frac1{2(2x_n+3)}<\frac32+\frac 12=2.$$
A: Induction hint:
$$
2 x_{n+1} = \dfrac{8+6x_n}{3+2x_n} = 3 - \dfrac{1}{3 + 2 x_n} \implies 2(x_{n+1} - x_n) = \dfrac{2(x_n - x_{n-1})}{(3+2x_n)(3+2x_{n-1})}
$$
The denominator is positive, so the difference between consecutive terms keeps the same sign.
