VMO

a) It is easy to see that $x_n>0$ for all $n\in {\mathbb{N}}^{\ast }$. For every positive integer $n$, we have $$\begin{array}{rl}|x_{n+1}-1|& =|\sqrt{x_n+8}-3+2-\sqrt{x_n+3}|\\ & =|(x_n-1)\left(\frac{1}{\sqrt{x_n+8}+3}-\frac{1}{\sqrt{x_n+3}+2}\right)|\\ & \le |x_n-1|\left(\frac{1}{\sqrt{x_n+8}+3}+\frac{1}{\sqrt{x_n+3}+2}\right)\\ & \le |x_n-1|\left(\frac{1}{3}+\frac{1}{2}\right)\\ & =\frac{5}{6}|x_n-1|.\end{array}$$ Therefore, $$|x_n-1|\le \frac{5}{6}|x_{n-1}-1|\le \cdots \le {\left(\frac{5}{6}\right)}^{n-1}|x_1-1|={\left(\frac{5}{6}\right)}^n,\forall n\in {\mathbb{N}}^{\ast }.$$ Note that $\lim {\left(\frac{5}{6}\right)}^n=0$, we obtain ${\lim }_{n\to \infty }{x}_n=1$.

b) Consider the function $$f(x)=\sqrt{x+8}-\sqrt{x+3}=\frac{5}{\sqrt{x+8}+\sqrt{x+3}},$$ with $x>0$, we see that $f(x)$ is a continuous and decreasing function on $(0,+\infty )$. Because $x_1>1$ then $x_2=f(x_1)<f(1)=1$, and $x_3=f(x_2)>f(1)=1,\dots$. In general, we can prove $x_{2k}<1<x_{2k-1}$ for all positive integers $k$.

Now, consider the function $g(x)=x+f(x)=x+\sqrt{x+8}-\sqrt{x+3}$ with $x>0$, we get $g(x)$ is a continuous function and $$g^{\prime }(x)=1+\frac{1}{2\sqrt{x+8}}-\frac{1}{2\sqrt{x+3}}>1-\frac{1}{2\sqrt{3}}>0,\forall x>0$$ so $g(x)$ is an increasing function on $(0,\infty )$. From here, we have the following claims If $x>1$ then $g(x)>g(1)=2$. If $0<x<1$ then $g(x)<g(1)=2$. Hence, $$x_{2k-1}+x_{2k}>2>x_{2k}+x_{2k+1},\forall k\in {\mathbb{N}}^{\ast }.$$ Now, we will prove the given inequality. Consider two cases: Case 1: $n=2k$ ($k\in {\mathbb{N}}^{\ast }$). It is easy to check that $2<x_1+x_2<3$ so the given inequality is true when $k=1$. Assume that $k>1$, we have $$\begin{gathered} x_1+x_2+\cdots +x_n=(x_1+x_2)+(x_3+x_4)+\cdots +(x_{2k-1}+x_{2k}) \\ >2+2+\cdots +2=2k \end{gathered}$$ and $$\begin{gathered} x_1+x_2+\cdots +x_n=x_1+(x_2+x_3)+\cdots +(x_{2k-2}+x_{2k-1})+x_{2k} \\ <2+2+\cdots +2+1=2k+1. \end{gathered}$$ Case 2: $n=2k-1$ ($k\in {\mathbb{N}}^{\ast }$). Clearly, the given inequality is true when $k=1$. Suppose that $k>1$, we have $$\begin{gathered} x_1+x_2+\cdots +x_n=(x_1+x_2)+\cdots +(x_{2k-3}+x_{2k-2})+x_{2k-1} \\ >2+2+\cdots +2+1=2k-1 \end{gathered}$$ and $$\begin{gathered} x_1+x_2+\cdots +x_n=x_1+(x_2+x_3)+\cdots +(x_{2k-2}+x_{2k-1}) \\ <2+2+\cdots +2=2k. \end{gathered}$$ To summarize, we have $n\le x_1+x_2+\cdots +x_n\le n+1$ for all positive integers $n$. $\square$