A two variable sequence 
I am having trouble with (a)
I am thinking about showing the following, $(x_{n+1},y_{n+1}) \geq (x_n,y_n)$ So being more explicit, I get
$\left (  \sqrt{\dfrac{x^2_n + 2y^2_n}{4}}, \dfrac{x_n + y_n + 1}{3}\right )\geq (x_n,y_n)$
So working out the first few steps of the algebra, I get
$$\left (  \dfrac{x^2_n + 2y^2_n}{4}, \dfrac{x_n + 1}{3}\right )\geq (x_n^2,\dfrac{2y_n}{3})$$
$$\left (  -3x^2_n + 2y^2_n, \dfrac{x_n + 1}{3}\right )\geq (0,\dfrac{2y_n}{3})$$
$$\left (  2y_n^2, -2y_n\right )\geq (3x_n^2,-(x_n + 1))$$
And now I am stuck...
 A: The proof of (a) is by induction. There is not quite increase always, since $x_1=0$.  But we can compute a step further, and show that $x_2\gt x_1$ and $y_2\gt y_1$.
Assume that $k \ge 1$, and $x_{k+1}\gt x_k$ and $y_{k+1}\gt y_k$. We prove that $x_{k+2}\gt y_{k+1}$ and $y_{k+2}\gt y_{k+1}$. 
This is relatively easy. For ease of typing, we deal with the $y$'s. 
We have $y_{k+2}=\frac{x_{k+1}+y_{k+1}+1}{3}$. But by the induction assumption, we have $x_{k+1}\gt x_k$ and $y_{k+1}\gt y_k$. It follows that
$$y_{k+2}=\frac{x_{k+1}+y_{k+1}}{3}\gt \frac{x_k+y_k+1}{3}=y_{k+1}.$$
The argument for $x_{k+2}$ is very similar.  
For boundedness, we prove by induction that $x_n \lt 1$ and $y_n \lt 1$ for all $n$. The two inuction steps are quite straightforward. 
Because our two sequences $(x_n)$ and $(y_n)$ are non-decreasing and bounded above, they each have a limit.
As to the limits, if $x$ is the limit of the $x_n$, and $y$ is the limit of the $y_n$, we have $x=\sqrt{\frac{x^2+2y^2}{4}}$ and $y=\frac{x+y+1}{3}$. If we square both sides of the first equation, we obtain an expression of $y$ in terms of $x$. Substituting for $x$ in the second equation, we find $y$. Then we compute $x$.
