We are given that , and

First we plug in into the recursions for and find that . Letting

and so on, we have and . Then we now want to prove that, for ,

First note that

(Cyclic relations hold as well.) Hence the original recursions can be rewritten as

\[ \begin{align*}\frac {x_n}{\sqrt {3n}} + \frac {n + 1}{3} = a_n & = a_{n - 1} + \frac {c_{n - 1}}{n} \\
& = \frac {x_{n - 1}}{\sqrt {3(n - 1)}} + \frac {n}{3} + \frac {z_{n - 1}}{n\sqrt {3(n - 1)}} + \frac {1}{3} \\
x_n & = \frac {x_{n - 1}\sqrt {n}}{\sqrt {n - 1}} + \frac {z_{n - 1}}{\sqrt {n(n - 1)}} \\
y_n & = \frac {y_{n - 1}\sqrt {n}}{\sqrt {n - 1}} + \frac {x_{n - 1}}{\sqrt {n(n - 1)}} \\
\quad z_n & = \frac {z_{n - 1}\sqrt {n}}{\sqrt {n - 1}} + \frac {y_{n - 1}}{\sqrt {n(n - 1)}}.\end{align*}\]

Summing the three new recurrence relations up, we have

But , so we can easily see that in general as well.

We now claim that , which would imply by the Cauchy-Schwarz Inequality that

or taking the square root (note that the right hand side parentheses contains a nonnegative expression) and using the triangle inequality,

\[ \begin{align*}4 > |x_n| + | - y_n| + | - z_n| & \ge|x_n - y_n - z_n| \\
& = |x_n + (x_n + y_n + z_n) - y_n - z_n| = |2x_n| \\
2 & > |x_n|,\]

as we need to prove for the original problem.

To prove this claim, notice that squaring our recursions for we have

\[ \begin{align*}x_n^2 & = \frac {n}{n - 1}x_{n - 1}^2 + \frac {1}{n(n - 1)}z_{n - 1}^2 + \frac {1}{n - 1}2x_{n - 1}z_{n - 1} \\
y_n^2 & = \frac {n}{n - 1}y_{n - 1}^2 + \frac {1}{n(n - 1)}x_{n - 1}^2 + \frac {1}{n - 1}2y_{n - 1}x_{n - 1} \\
z_n^2 & = \frac {n}{n - 1}z_{n - 1}^2 + \frac {1}{n(n - 1)}y_{n - 1}^2 + \frac {1}{n - 1}2z_{n - 1}y_{n - 1}.\end{align*}\]

Let , so . Adding these three squared quantities together and using

\[ \begin{align*} & 2x_{n - 1}y_{n - 1} + 2y_{n - 1}z_{n - 1} + 2z_{n - 1}x_{n - 1} \\
= & (x_{n - 1} + y_{n - 1} + z_{n - 1})^2 - (x_{n - 1}^2 + y_{n - 1}^2 + z_{n - 1}^2) \\
= & (0)^2 - s_{n - 1} = - s_{n - 1}\end{align*}\]

we have

\[ \begin{align*}s_n & = x_n^2 + y_n^2 + z_n^2 = \left(\frac {n}{n - 1} + \frac {1}{n(n - 1)}\right)s_{n - 1} + \frac1{n - 1}( - s_{n - 1}) \\
& = \left(\frac {n}{n - 1} + \frac {1}{n(n - 1)} - \frac {1}{n - 1}\right)s_{n - 1} = \left(1 + \frac {1}{n(n - 1)}\right)s_{n - 1} \\
& = \left(1 + \frac {1}{n(n - 1)}\right)\cdots\left(1 + \frac {1}{2\cdot1}\right)s_{1} \\
& = 2\left(1 + \frac {1}{n(n - 1)}\right)\cdots\left(1 + \frac {1}{2\cdot1}\right).\end{align*}\]

Therefore it is sufficient to prove that for all positive (when , ),

\[ \begin{align*}\frac {16}{3} > s_{n + 1} & = 2\left(1 + \frac1{1\cdot2}\right)\cdots\left(1 + \frac {1}{n(n + 1)}\right) \\
& = 2\prod_{k = 1}^{n}\left(1 + \frac {1}{k(k + 1)}\right) \\
& = 2\prod_{k = 1}^{n}\frac {k^2 + k + 1}{k^2 + k},\end{align*}\]

or dividing both sides by and taking the reciprocal (reversing the direction of the inequality),

Since , we can calculate for from the recursion that and , both of which are less than . Otherwise, , so because is true for all positive we have

\[ \begin{align*} & \prod_{k = 1}^{n}\left(1 - \frac {1}{k^2 + k + 1}\right) \\
> & \left(1 - \frac {1}{1^2 + 1 + 1}\right)\left(1 - \frac {1}{2^2 + 2 + 1}\right)\prod_{k = 3}^{n}\left(1 - \frac {1}{k^2}\right) \\
= & \left(\frac {2}{3}\right)\left(\frac {6}{7}\right)\prod_{k = 3}^{n}\left(\left(1 - \frac {1}{k}\right)\left(1 + \frac {1}{k}\right)\right) = \frac {4}{7}\prod_{k = 3}^{n}\frac {k - 1}{k}\prod_{k = 3}^{n}\frac {k + 1}{k} \\
= & \frac {4}{7}\left(\frac {\prod_{k = 3}^{n}(k - 1)\prod_{k = 3}^{n}(k + 1)}{\prod_{k = 3}^{n}k\prod_{k = 3}^{n}k}\right) = \frac {4}{7}\left(\frac {\left(\frac {2}{n}\prod_{k = 3}^{n}k\right)\left(\frac {n + 1}{3}\prod_{k = 3}^{n}k\right)}{\prod_{k = 3}^{n}k\prod_{k = 3}^{n}k}\right) \\
= & \frac {4}{7}\left(\frac {2}{3}\right)\left(\frac {n + 1}{n}\right) = \frac {8}{21}\left(\frac {n + 1}{n}\right) = \left(\frac {3}{8} + \frac {1}{168}\right)\left(1 + \frac {1}{n}\right) > \frac {3}{8}.\blacksquare\end{align*}\]