Boundedness of solutions of a difference equation Is there someone who can show me how I can prove this conjecture? Or at least show me how to do the first implication ? 

Conjecture: Assume $\alpha,\beta, \lambda \in [0,\infty)$. Then every positive solution of the difference equation : 
  $$z_{n+1}=\frac{\alpha+z_{n}\beta +z_{n-1}\lambda}{z_{n-2}},\quad n=0,1,\ldots$$ 
  is bounded if and only if $\beta=\lambda$

Any help is very welcome. Thank you for any comments or any replies.
Edit: as mentioned in the comments, this is conjecture 8 in this paper by Ladas, Lugo and Palladino
 A: Here is just an idea, which may or may not work. Suppose that $\beta=\lambda>0$. Let $t_n:=z_n/\beta$ and $c:=\alpha/\beta^2$. Then the dynamics can be rewritten as 
$$(!)\qquad t_{n+1}=\frac{c+t_n+t_{n-1}}{t_{n-2}}
$$
(say for $n=2,3,\dots$), just with one parameter $c\ge0$. To prove the "if" part of the conjecture, it would be enough to construct, for each nonnegative $c$, a "sub-energy" function $f_c\colon(0,\infty)^3\to\mathbb{R}$ such that 
$$(!!)\qquad f_c(t_0,t_1,t_2)\to\infty\quad\text{as}\quad t_0+t_1+t_2\to\infty$$ 
and 
for some natural $k$ and all $t=(t_0,t_1,t_2)\in(0,\infty)^3$ one has the "sub-energy" inequality
$f_c(T^k t)\le f_c(t)$, where $Tt:=(t_1,t_2,t_3)$, with $t_3=\frac{c+t_2+t_1}{t_0}$, according to the dynamics. Of course, $T^k$ is the $k$th power of the operator $T$. For $k=1$, the sub-energy inequality is the functional inequality 
$$(*)\qquad f_c\Big(t_1,t_2,\frac{c+t_2+t_1}{t_0}\Big)\le f_c(t_0,t_1,t_2) \quad
\text{for all positive $t_0,t_1,t_2$, }
$$
with an unknown function $f_c$. 
To construct a sub-energy function, one might want to start with some easy function $f_{c,0}$ such that $f_{c,0}(t_0,t_1,t_2)\to\infty$ as $t_0+t_1+t_2\to\infty$, and then consider something like $f_{c,0}\vee(f_{c,0}\circ T^k)\vee(f_{c,0}\circ T^{2k})\vee\dots$. 
Perhaps similar ideas could also work for the "only if" part.
Addendum: Inequality $(*)$ can be obviously restated in the following more symmetric form: 
$$(**)\qquad 
t_0t_3=c+t_1+t_2\implies f_c(t_1,t_2,t_3)\le f_c(t_0,t_1,t_2)  
$$
for all positive real $t_0,t_1,t_2,t_3$.  
Addendum 2: The condition $t_0+t_1+t_2\to\infty$ in $(!!)$ can be replaced by any one of the following (stronger) conditions: (i) $t_0\to\infty$ or (ii) $t_1\to\infty$ or (iii) $t_2\to\infty$; this of course will replace condition $(!!)$ by a weaker condition, which will make it easier to construct a sub-energy function $f_c$. 
Here are details: Suppose that $(*)$ holds for some function $f_c$ such that $f_c(t_0,t_1,t_2)\to\infty$ as $t_0\to\infty$. Suppose that, nonetheless, a positive sequence $(t_0,t_1,\dots)$ satisfying condition $(!)$ is unbounded, so that, as $k\to\infty$, one has $t_{n_k}\to\infty$ for some sequence $(n_k)$ of natural numbers. Then $f_c(t_{n_k},t_{1+n_k},t_{2+n_k})\to\infty$ as $k\to\infty$. This contradicts $(*)$, which implies, by induction, that $f_c(t_n,t_{1+n},t_{2+n})\le f_c(t_0,t_1,t_2)$ for all natural $n$. Quite similarly one can do with (ii) $t_1\to\infty$ or (iii) $t_2\to\infty$ in place of (i) $t_0\to\infty$. 
Also, instead of the dynamics of the triples $(t_n,t_{1+n},t_{2+n})$ one can consider the corresponding dynamics (in $n$) of the consecutive $m$-tuples $(t_n,\dots,t_{m-1+n})$ for any fixed natural $m$. 
Also, instead of inequality $f_c(t_1,t_2,t_3)\le f_c(t_0,t_1,t_2)$ in $(*)$, one may consider a weaker inequality like $f_c(t_2,t_3,t_4)\le f_c(t_0,t_1,t_2)\vee f_c(t_1,t_2,t_3)$ for all positive $t_0,\dots,t_4$ satisfying condition $(!)$. 
Addendum 3: One can try to do the "only if" part in a similar manner. Suppose that $0<\beta\ne\lambda>0$. Let $u_n:=z_n/\sqrt{\beta\lambda}$, $c:=\alpha/(\beta\lambda)$, and $a:=\sqrt{\beta/\lambda}\ne1$. Then the dynamics can be rewritten as 
$$(!!!)\qquad u_{n+1}=\frac{c+au_n+u_{n-1}/a}{u_{n-2}}, 
$$
just with two parameters, $c\ge0$ and $a>0$. Suppose one can construct, for each pair $(c,a)\in[0,\infty)\times\big((0,\infty)\setminus\{1\}\big)$ and some $\rho=\rho_{c,a}\in(1,\infty)$, a "$\rho$-super-energy" function $g=g_{a,c;\rho}\colon(0,\infty)^3\to(0,\infty)$ such that $g$ is bounded on each bounded subset of $(0,\infty)^3$ and 
$$(***)\qquad g\Big(u_1,u_2,\frac{c+au_2+u_1/a}{u_0}\Big)\ge\rho\, g(u_0,u_1,u_2)\quad 
\text{for all positive $u_0,u_1,u_2$.}
$$
Then, by induction, $g(u_n,u_{1+n},u_{2+n})\ge\rho^n g(u_0,u_1,u_2)\to\infty$ as $n\to\infty$, for any sequence $(u_n)$ satisfying $(!!!)$. Therefore and because $g$ is bounded on each bounded subset of $(0,\infty)^3$, it would follow that the sequence $(u_n)$ is unbounded. 
Addendum 3a: For any pair $(c,a)\in[0,\infty)\times(0,\infty))$ and any $\rho\in(1,\infty)$, there is no "$\rho$-super-energy" function $g\colon(0,\infty)^3\to(0,\infty)$. This follows because the point $(u_{a,c},u_{a,c},u_{a,c})$ with 
$u_{a,c}:=\dfrac{1+a^2+\sqrt{a^4+4 a^2 c+2 a^2+1}}{2 a}$ is a fixed point (in fact, the only fixed point) of the map $T$ given by the formula $T(u_0,u_1,u_2)=\Big(u_1,u_2,\dfrac{c+au_2+u_1/a}{u_0}\Big)$. (If $a\ne1$, then this point is the only fixed point of the map $T^2$ as well.)
This also disproves, in general, the "only if" part of the conjecture in question. 
However, one may now try to amend this conjecture by excluding the initial point $(u_{a,c},u_{a,c},u_{a,c})$. Then, accordingly, the definition of a "$\rho$-super-energy" function would have it defined on a subset (say $S$) of the set $(0,\infty)^3\setminus\{(u_{a,c},u_{a,c},u_{a,c})\}$, instead of $(0,\infty)^3$; such a subset may be allowed to depend on the choice of the initial point $(u_0,u_1,u_2)$, say on its distance from the fixed point $(u_{a,c},u_{a,c},u_{a,c})$, and one would then have to also prove that $S$ is invariant under the map $T$.   
A: Note :This is comments for the answer related to the question and not lead to the Answer 
of the above question  .and may it is helpful to be constrictive for a complet proof 
The discussion of energy functions is at best very difficult to understand and at worst incorrect or incomplete. To illustrate this consider that the Answer above  emphasizes that it is enough for $f(x,y,z)$ to go to infinity as x goes to infinity. This of course is patently false for a general system, (consider $x_{n}=5x_{n-1}, y_{n}=2y_{n-1}, z_{n}=2z_{n-1}$, and $f(x,y,z)=x)$, and only true for a delay dynamical system in one dimension. This was perhaps to be assumed by context, but I didn't see it mentioned. Moreover the conjecture  wasn't proved.
"In a short statement":The proof is not standard and the theory of Lyapunouv functions cannot solves this type of problems. 
