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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 $a:=\sqrt{\beta/\lambda}$. 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 $(c,a)\in[0,\infty)\times(0,\infty)$ 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 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 $T$.

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 $(!)$.

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 $a:=\sqrt{\beta/\lambda}$. 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 $(c,a)\in[0,\infty)\times(0,\infty)$ 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.

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 $$t_{n+1}=\frac{c+t_n+t_{n-1}}{t_{n-2}} $$ (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 $f_c(t_0,t_1,t_2)\to\infty$ as $t_0+t_1+t_2\to\infty$ 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 $$(*)\qquad f_c\Big(t_1,t_2,\frac{c+t_2+t_1}{t_0}\Big)\le f_c(t_0,t_1,t_2) $$ 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 $t_0,t_1,t_2,t_3$.

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 $(!)$.