China Western Mathematical Olympiad

(1) When $b<0$ and $m$ is even, in order that $a{b}^{m-1}<-2$, we should first have $a{b}^m+b>-b>0$, and therefore $a(a{b}^m+b{)}^m+b>a{b}^m+b>0$, i.e. $x_3>x_2>0$. Using the fact that $a{x}^m+b$ is monotonically increasing on $(0,+\infty )$, it can be established that each succeeding term of the sequence $\{x_n\}$ is greater than its preceding term, and is greater than $-b$ starting from the second term.

Considering any three consecutive terms of the sequence $x_n,x_{n+1},x_{n+2},\dots$, we have $$\begin{array}{rl}{x}_{n+2}-x_{n+1}& =a(x_{n+1}^m-x_n^m)\\ & =a(x_{n+1}-x_n)(x_{n+1}^{m-1}+x_{n+1}^{m-2}{x}_n+\cdots +x_n^{m-1})\\ & >am{x}_n^{m-1}(x_{n+1}-x_n)\\ & >am(-b{)}^{m-1}(x_{n+1}-x_n)\\ & >2m(x_{n+1}-x_n)\\ & >x_{n+1}-x_n.\end{array}$$ It is obvious that the difference of any two consecutive terms of the sequence $\{x_n\}$ is increasing, and hence it is not bounded.

When $a{b}^{m-1}\ge -2$, mathematical induction is used to prove that each term of the sequence $\{x_n\}$ falls on the interval $[b,-b]$.

The first term $b$ falls on the interval $[b,-b]$. Suppose that the term $x_n$ satisfies the condition $b\le x_n\le -b$ for a particular $n$. Then $0\le x_n^m\le b^m$, and hence $$b=a\times 0^m+b\le x_{n+1}\le a{b}^m+b\le -b.$$ Thus, the sequence $\{x_n\}$ is bounded if and only if $a{b}^{m-1}\ge -2$.

(2) When $b>0$, each term of the sequence $\{x_n\}$ is positive. So, we first prove that $\{x_n\}$ is bounded if and only if the equation $a{x}^m+b=x$ has positive real roots.

Suppose that $a{x}^m+b=x$ has no positive real roots. In such a case, the minimum value of the function $p(x)=a{x}^m+b-x$ on the interval $(0,+\infty )$ is greater than zero. Let $t$ be the minimum value. It follows that for any two consecutive terms of the sequence $x_n$ and $x_{n+1}$, we have $x_{n+1}-x_n=a{x}_n^m-x_n+b$. Thus, each succeeding term of the sequence $\{x_n\}$ is greater than the preceding term by at least $t$. Hence, it is not bounded.

If the equation $a{x}^m+b=x$ has positive real roots, let $x_0$ be one of the positive real roots. Then, by using mathematical induction, we prove that each term of the sequence $\{x_n\}$ is less than $x_0$. Firstly, the first term $b$ is less than $x_0$. Suppose that $x_n<x_0$ for a particular $n$. By virtue of the fact that $a{x}^m+b$ is increasing on the interval $[0,+\infty )$, it can be established that $$x_{n+1}=a{x}_n^m+b<a{x}_0^m+b=x_0.$$ Therefore, the sequence is bounded.

Further, the equation $a{x}^m+b=x$ has positive roots if and only if the minimum value of $a{x}^{m-1}+\frac{b}{x}$ on the interval $(0,+\infty )$ is not greater than $1$, whereas the minimum value of $a{x}^{m-1}+\frac{b}{x}$ can be determined by mean inequality, i.e. $$a{x}^{m-1}+\frac{b}{x}=a{x}^{m-1}+\frac{b}{(m-1)x}+\cdots +\frac{b}{(m-1)x}\ge m\sqrt[m]{\frac{a{b}^{m-1}}{(m-1{)}^{m-1}}}$$ As such, the sequence $\{x_n\}$ is bounded if and only if $$m\sqrt[m]{\frac{a{b}^{m-1}}{(m-1{)}^{m-1}}}\le 1,\text{ i.e. }a{b}^{m-1}\le \frac{(m-1{)}^{m-1}}{m^m}.$$

When $b<0$, and $m$ is odd, let $y_n=-x_n$. Then $y_1=-b>0$, $y_{n+1}=a{y}_n^m+(-b)$, showing that the sequence $\{x_n\}$ is bounded if and only if the sequence $\{y_n\}$ is bounded. Thus, by using the above reasoning, it can be proven that (2) holds.