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$x_n$ eventually decreases.

Denote $\beta=\sqrt\alpha$, then $0<\beta<1/2$. Then $x_n$ is the least positive number for which $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\geqslant x, \, \forall x\in [0,x_n].\tag{1}$$ Here $[t^a]F(t)$ denotes a coefficient of $t^a$ in the Laurent polynomial $F$.

So, in order to prove that $x_n\leqslant x_{n-1}$, it suffices to prove that (1) holds with exponent $n-2$ instead $n-1$, that in turns follows from the inequality $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\leqslant [t^0](\beta(t+t^{-1})x+1-x)^{n-2}\tag{2} $$ for all $x\in [0,x_n]$. (2) may be proved by applying the formula $$[t^0]A(t)=\frac1{2\pi}\int_{0}^{2\pi}A(e^{is})ds$$ for any Laurent polynomial $A(t)$ and in our situation $\beta(e^{is}+e^{-is})x+1-x=1-x(1-2\beta \cos s)\leqslant 1$ and also $1-x(1-2\beta \cos s)\geqslant 0$ provided that $x_n\leqslant 1/(1+2\beta)$, that holds for large enough $n$ as shown by fedja

$x_n$ eventually decreases.

Denote $\beta=\sqrt\alpha$, then $0<\beta<1/2$. Then $f_n(0)=1>0$ and therefore $x_n$ is the maximal positive number for which $$f_n(x)=[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\geqslant x, \, \forall x\in [0,x_n].\tag{1}$$ Here $[t^a]F(t)$ denotes a coefficient of $t^a$ in the Laurent polynomial $F$.

So, in order to prove that $x_n\leqslant x_{n-1}$, it suffices to prove that (1) holds with exponent $n-2$ instead $n-1$, that in turns follows from the inequality $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\leqslant [t^0](\beta(t+t^{-1})x+1-x)^{n-2}\tag{2} $$ for all $x\in [0,x_n]$. (2) may be proved by applying the formula $$[t^0]A(t)=\frac1{2\pi}\int_{0}^{2\pi}A(e^{is})ds$$ for any Laurent polynomial $A(t)$ and in our situation $\beta(e^{is}+e^{-is})x+1-x=1-x(1-2\beta \cos s)\leqslant 1$ and also $1-x(1-2\beta \cos s)\geqslant 0$ provided that $x_n\leqslant 1/(1+2\beta)$, that holds for large enough $n$, as shown by fedja.

Eventually monotonicity.

Denote $\beta=\sqrt\alpha$, then $0<\beta<1/2$. Then $x_n$ is the least positive number for which $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\geqslant x, \, \forall x\in [0,x_n].\tag{1}$$ So, in order to prove that $x_n\leqslant x_{n-1}$, it suffices to prove that (1) holds with exponent $n-2$ instead $n-1$, that in turns follows from the inequality $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\leqslant [t^0](\beta(t+t^{-1})x+1-x)^{n-2}\tag{2} $$ for all $x\in [0,x_n]$. (2) may be proved by applying the formula $$[t^0]A(t)=\frac1{2\pi}\int_{0}^{2\pi}A(e^{is})ds$$ for any Laurent polynomial $A(t)$ and in our situation $\beta(e^{is}+e^{-is})x+1-x=1-x(1-2\beta \cos s)\leqslant 1$ and also $1-x(1-2\beta \cos s)\geqslant 0$ provided that $x_n\leqslant 1/(1+2\beta)$

$x_n$ eventually decreases.

Denote $\beta=\sqrt\alpha$, then $0<\beta<1/2$. Then $x_n$ is the least positive number for which $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\geqslant x, \, \forall x\in [0,x_n].\tag{1}$$ Here $[t^a]F(t)$ denotes a coefficient of $t^a$ in the Laurent polynomial $F$.

So, in order to prove that $x_n\leqslant x_{n-1}$, it suffices to prove that (1) holds with exponent $n-2$ instead $n-1$, that in turns follows from the inequality $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\leqslant [t^0](\beta(t+t^{-1})x+1-x)^{n-2}\tag{2} $$ for all $x\in [0,x_n]$. (2) may be proved by applying the formula $$[t^0]A(t)=\frac1{2\pi}\int_{0}^{2\pi}A(e^{is})ds$$ for any Laurent polynomial $A(t)$ and in our situation $\beta(e^{is}+e^{-is})x+1-x=1-x(1-2\beta \cos s)\leqslant 1$ and also $1-x(1-2\beta \cos s)\geqslant 0$ provided that $x_n\leqslant 1/(1+2\beta)$, that holds for large enough $n$ as shown by fedja

Eventually monotonicity.

Denote $\beta=\sqrt\alpha$, then $0<\beta<1/2$. Then $x_n$ is the least positive number for which $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\geqslant x, \, \forall x\in [0,x_n].\tag{1}$$ So, in order to prove that $x_n\geqslant x_{n-1}$, it suffices to prove that (1) holds with exponent $n-2$ instead $n-1$, that in turns follows from the inequality $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\leqslant [t^0](\beta(t+t^{-1})x+1-x)^{n-2}\tag{2} $$ for all $x\in [0,x_n]$. (2) may be proved by applying the formula $$[t^0]A(t)=\frac1{2\pi}\int_{0}^{2\pi}A(e^{is})ds$$ for any Laurent polynomial $A(t)$ and in our situation $\beta(e^{is}+e^{-is})x+1-x=1-x(1-2\beta \cos s)\leqslant 1$ and also $1-x(1-2\beta \cos s)\geqslant 0$ provided that $x_n\leqslant 1/(1+2\beta)$

Eventually monotonicity.

Denote $\beta=\sqrt\alpha$, then $0<\beta<1/2$. Then $x_n$ is the least positive number for which $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\geqslant x, \, \forall x\in [0,x_n].\tag{1}$$ So, in order to prove that $x_n\leqslant x_{n-1}$, it suffices to prove that (1) holds with exponent $n-2$ instead $n-1$, that in turns follows from the inequality $$[t^0](\beta(t+t^{-1})x+1-x)^{n-1}\leqslant [t^0](\beta(t+t^{-1})x+1-x)^{n-2}\tag{2} $$ for all $x\in [0,x_n]$. (2) may be proved by applying the formula $$[t^0]A(t)=\frac1{2\pi}\int_{0}^{2\pi}A(e^{is})ds$$ for any Laurent polynomial $A(t)$ and in our situation $\beta(e^{is}+e^{-is})x+1-x=1-x(1-2\beta \cos s)\leqslant 1$ and also $1-x(1-2\beta \cos s)\geqslant 0$ provided that $x_n\leqslant 1/(1+2\beta)$