Nested Sequences and Boundeness For three sequences $(x_n), (y_n), (z_n)$ with positive starting elements $x_1, y_1, z_1$, we have the following relationship:
$$ x_{n+1} = y_n + \frac{1}{z_n} \quad y_{n+1} = z_n + \frac{1}{x_n} \quad z_{n+1} = x_n + \frac{1}{y_n} \quad (n = 1,2,3, \ldots)$$
(a) Prove that none of the three sequences is bounded from above.
(b) Prove that at least one of the numbers $x_{200}, y_{200}, z_{200}$ is greater than $20$.
I solved (a). Using contradiction and the relationship
$$x_{n+1}+y_{n+1}+z_{n+1}=x_1+y_1+z_1+\sum_{k=1}^{n-1}\frac{1}{x_k}+\frac{1}{y_k}+\frac{1}{z_k}.$$
I tried to use similar method to solve (b) but I could not, can anyone help? Thanks.
 A: Let $(a_n)_{n\geq 2}$ be the sequence defined by
$a_2=2$ and $a_{n+1}=a_n+\frac{1}{a_n}$.
Lemma. For every $n\geq 2$, at least one of $x_n,y_n,z_n$
is $\geq a_n$.
Proof of lemma: For $n=2$, if we put $f(t)=t+\frac{1}{t}$ we have
$f(t)\geq 2$  and $x_2+y_2+z_2=f(x_1)+f(y_1)+f(z_1) \geq 6$, so one
of $x_2,y_2,z_2$ is indeed $\geq a_2$.
  Next, let us argue by induction for $n>2$. So suppose (induction
  hypothesis) that at least one of $x_m,y_m,z_m$
is $\geq a_m$, where $m=n-1$. By symmetry, we may assume
$x_m \geq a_m$. 
If $y_m\leq a_m$, then $z_n=x_m+\frac{1}{y_m} \geq a_n$ and we are done.
So assume $y_m\geq a_m$.
If $z_m\leq a_m$, then $x_n=y_m+\frac{1}{z_m} \geq a_n$ and we are done.
So assume $z_m\geq a_m$.
Since all the numbers $x_m,y_m,z_m$ are $\geq a_m$. Since $f$
is increasing on $(2,\infty)$, all the numbers $f(x_m),f(y_m),f(z_m)$
are $\geq f(a_m)=a_n$. So $x_n+y_n+z_n=f(x_m)+f(y_m)+f(z_m) \geq 3a_m$
and at least one of $x_n,y_n,z_n$ must be $\geq a_n$. This concludes the proof of the lemma.
Now, since $a_{k+1}^2-a_k^2=2+\frac{1}{a_k^2} \geq 2$, we deduce
$$
a_{200}^2=a_2^2+\sum_{k=2}^{199} a_{k+1}^2-a_k^2
\geq 4+2\times 198=400
$$
whence $a_{200} \geq 20$ as wished.
A: since use AM-GM inequality
$$x^2_{n+1}+y^2_{n+1}+z^2_{n+1}=x^2_{n}+y^2_{n}+z^2_{n}+2\left(\dfrac{y_{n}}{z_{n}}+\dfrac{z_{n}}{x_{n}}+\dfrac{x_{n}}{y_{n}}\right)+\dfrac{1}{x^2_{n}}+\dfrac{1}{y^2_{n}}+\dfrac{1}{z^2_{n}}\ge x^2_{n}+y^2_{n}+z^2_{n}+2\cdot 3$$
so
$$x^2_{200}+y^2_{200}+z^2_{200}>x^2_{2}+y^2_{2}+z^2_{2}+6\cdot 198$$
and use AM-GM inequality
$$x^2_{2}+y^2_{2}+z^2_{2}=\left(x^2_{1}+\dfrac{1}{x^2_{1}}\right)+2\left(\dfrac{y_{1}}{z_{1}}+\dfrac{z_{1}}{x_{1}}+\dfrac{x_{1}}{y_{1}}\right)\ge 12$$
so
$$x^2_{200}+y^2_{200}+z^2_{200}>6\times 200$$
so
$$\max{(x^2_{200},y^2_{200},z^2_{200})}>400$$
