The sequence $x_1, x_2, x_3, \ldots$ is such that $x_1 = 1 $ and
$$x_{n+1} \space = \frac{1+4x_n}{5 + 2x_n}$$
Prove by induction that $x_n > 0.5$ for all $n \ge 1$.
I have absolutely no clue how to go about this one. Can someone please explain.
Very sorry about the error, that's 0.5 and not 1.
As others have commented, the problem as stated is (twice) incorrect, since $x_1 = 1$ and $x_2 = 5/7 < 1$, and it is therefore impossible to give an answer as is.
However, it is possible to make a complete study of the sequence $x_n$ in the following way: it is easy to see that if we define two sequences $y_n, z_n$ by $$ y_1 = x_1, z_1 = 1, \quad y_{n+1} = 4 y_n + z_n, \quad z_{n+1} = 2 y_n + 5 z_n,$$ then we shall always have $x_n = y_n/z_n$. This is a linear recurrence with characteristic polynomial $r^2 - (4+5) r + (4\times 5-2\times 1) = r^2 - 9 r + 18 = (r-6) (r-3)$, which therefore has the characteristic roots $3$ and $6$. So we write $y_n$ and $z_n$ as linear combinations of $3^n$ and $6^n$: we find $$y_n = \frac{3^n+6^n)}{9},\quad z_n = \frac{-3^n+2\cdot 6^n}{9},$$ (manually check that the recurrence still holds!), so that $$x_n = \frac{3^n+6^n}{-3^n+2\cdot 6^n} = \frac{1+2^n}{-1+2\cdot 2^n}$$. (Again, manually check that this is actually $x_n$!). This means that
- the values taken by the sequence $(x_n)$ lie in the image by the function $t \mapsto (1+t)/(-1+2t)$ of the interval $[1,+\infty[$;
- the sequence $x_n$ converges to $1/2$.